Physics M.C.Q (1 Marks)

$V_3>V_2>V_1$

$\Rightarrow P_1>P_2>P_3$

(We draw a straight line parallel to volume axis to get this)

$\frac{ v _1}{ v _2}=\sqrt{\frac{ T _1}{ T _2}}$

$=$ let initial speed is $v$

As speed is increased by $3$ times so final speed become $4 v$

$\Rightarrow \frac{v}{4 v}=\sqrt{\frac{223}{T}}$

$T=3568\,K$

So temp. in ${ }^{\circ} C =3568-273=3295^{\circ}\,C$

$=\frac{\text { mass }}{\text { molar mass }}\left(22.4 \times 10^{-3} m ^{3}\right)$

$=\frac{4.5 \times 10^{3}}{18} \times 22.4 \times 10^{-3} m ^{3}$

$=\text { maximum for } He$

$(B)$ $P=\frac{1}{3} n m V_{m}^{2}$

$(C)$ $\mathrm{E}=\frac{3}{2} \mathrm{kT}$

$(D)$ $E_{\text {Total }}=\mathrm{n} \frac{\mathrm{f}}{2} \mathrm{R} T=\frac{5}{2} \mathrm{RT}$

$\frac{P}{\rho}=\frac{R T}{M w}$

$\rho=\frac{ PMw }{ RT }=\frac{249 \times 10^{3} \times 2 \times 10^{-3}}{8.314 \times 300}=0.199$

$\rho=0.2\, kg / m ^{3}$

$PV =\mu RT$

$PV =\frac{ M }{ M _{0}} RT$

$P =\left(\frac{M}{V}\right) \frac{R T}{M_{0}}$

$P=\frac{\rho R T}{M_{0}}$

where $\rho$ is mass density and $M _{0}$ is molar mass.

where $3$ is translational degree of freedom For monoatomic gas total degree of freedom

$f =3$ (translational degree of freedom)

$\lambda_{ m }=\frac{1}{\sqrt{2} \pi d ^{2} n }$

where $d$ is diameter of a gas molecule and $n$ is molecular density

$\ell=\frac{1}{\sqrt{2} n \pi d^{2}}$

$\ell \propto \frac{1}{d^{2}}$

As temperature increases $\mathrm{KE}$ also increases

$v_{\text {escape }}=11200 \mathrm{ms}^{-1}$

Say at temperature $T,$ oxygen molecule attains escape velocity.

So, $v_{\text {escape }}=\sqrt{\frac{3 k_{B} T}{m_{\mathrm{O}_{2}}}} \Rightarrow 11200=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}}$

On solving, $T=8.360 \times 10^{4} \mathrm{K}$

$V_{r m s} \propto \sqrt{T}$

$\frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{303}{363}}$

$\frac{V_1}{V_2}=1.1$

$\frac{V_2}{V_1}-1=1.1-1 \quad[\because$ Subtract 1 from both sides $]$

$\frac{\Delta V}{V} \times 100$

Percentage incrase $\frac{\Delta V}{V}=10 \%$

                                                                    $. . .(i)$

Density, $\rho=\frac{\text { mass }}{\text { volume }}=\frac{(\text { molar mass }) P}{R T}=\frac{\left(m N_{A}\right) P}{R T}$

                                                           $[\text { From eqn. (i) }]$

$\rho=\frac{m P}{k T}$                    $\left(\because R=N_{A} k\right)$

${\therefore \frac{v_{27}}{v_{127}}=\sqrt{\frac{27+273}{127+273}}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2}} $

${\text { or } v_{127}=\frac{2}{\sqrt{3}} \times v_{27}=\frac{2}{\sqrt{3}} \times 200 \mathrm{m} \mathrm{s}^{-1}=\frac{400}{\sqrt{3}} \mathrm{m} \mathrm{s}^{-1}}$

$PV^3=constant$

For a Polytropic process, $PV^\alpha=constant$

 $C = {C_v} + \frac{R}{{1 - \alpha }} = \frac{3}{2}R + \frac{R}{{1 - 3}} = R$

Hence, final common temperature of the system will be closer to $100^{\circ}\,C$.

$\therefore T _{ c }=50^{\circ}\,C$

The line $l$ corresponds to a liquid line saturated. The line $g$ corresponds to a saturated gas line:

The vegion between is the wet region (both lignid and gas). $T_{c}$ corresponds to the critical temperature.

So, it is clear that the temp must be below the critical temperature for a matler to exist simultaneously as gas and liquid.

Correct choice - option-$C$

$\therefore$ $\frac{{{\rho _1}}}{{{\rho _2}}} = \frac{{{P_1}}}{{{P_2}}} \times \frac{{{T_2}}}{{{T_1}}}$

$\frac{{{\rho _{top}}}}{{{\rho _{bottom}}}} = \frac{{{P_{top}}}}{{{P_{bottom}}}} \times \frac{{{T_{bottom}}}}{{{T_{top}}}} = \frac{{70}}{{76}} \times \frac{{300}}{{280}} = \frac{{75}}{{76}}$

Bubble rises to height, $d=40 m$

Temperature at a depth of $40 m , T_{1}=12^{\circ} C =285 K$

Temperature at the surface of the lake, $T_{2}=35^{\circ} C =308 K$

The pressure on the surface of the lake:

$P_{2}=1 atm =1 \times 1.013 \times 10^{5} Pa$

The pressure at the depth of $40 m$

$P_{1}=1 atm +d\rho g$ Where $, \rho$ is the density of

water $=10^{3} kg / m ^{3} g$ is the acceleration due

to gravity $=9.8 m / s ^{2}$

$\therefore P_{1}=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8=493300 Pa$

We have: $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

Where, $V_{2}$ is the volume of the air bubble when it reaches the surface $V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$

$=\frac{(493300)\left(1.0 \times 10^{-6}\right) 308}{285 \times 1.013 \times 10^{5}}$

$=5.263 \times 10^{-6}\; m ^{3}$ or $5.263 \;cm ^{3}$

Therefore, when the air bubble reaches the surface, its volume becomes $5.263 \;cm ^{3} .$

Let there are $n_1$ moles of hydrogen and $n_2$ moles of helium in the given mixture. As $Pv = nRT$

Then the pressure of the mixture

$P =\frac{ n _1 RT }{ V }+\frac{ n _2 RT }{ V }=\left( n _1+ n _2\right) \frac{ RT }{ V }$

$\Rightarrow 2 \times 101.3 \times 10^3=\left( n _1+ n _2\right) \times \frac{(8.3 \times 300)}{20 \times 10^{-3}}$

$\text { or, } \left( n _1+ n _2\right)=\frac{2 \times 101.3 \times 10^3 \times 20 \times 10^{-3}}{(8.3)(300)}$

$\text { or, } \quad n _1+ n _2=1.62$

The mass of the mixture is (in grams) $n _1 \times 2+ n _2 \times 4=5$

$\Rightarrow \quad\left(n_1+2 n_2\right)=2.5 \quad \ldots(2)$

Solving the eqns. (1) and (2), we get $n _1=0.74$ and $n _2=0.88$

Hence, $\frac{ m _{ H }}{ m _{ He }}=\frac{0.74 \times 2}{0.88 \times 4}=\frac{1.48}{3.52}=\frac{2}{5}$

or ${P_2} = \frac{{{P_1}{T_2}}}{{{T_1}}}$ $ = \frac{{P(273 + 927)}}{{(273 + 27)}}$ $=4P$

$\Rightarrow {T_2} = 2 \times {T_1}$ $ = 2 \times (273 + 0) = 546K$

$\Rightarrow$  ${T_2} = 273 \times 2 = 546K$

$\Rightarrow 273^\circ C$ $\Rightarrow 273$ $^\circ$$ C$

$\Rightarrow$  ${T_2} = 900K$ $⇒$ $627°C$       $[ T(K) = 273 + t°C]$

The density of a gas at normal pressure and $27^{\circ} C$ or $300 K$ temperature $=24$

then the density at $127^{\circ} C$ on $400 K$, pressure is constant,

So. We know,

$d=\frac{P M}{R T}$

$d \propto \frac{1}{T}$

$\Rightarrow  \frac{d_{1}}{d_{2}} =\frac{T_{2}}{T_{1}}$

$\Rightarrow  d_{2} =\frac{24 \times 300}{400}$

$d_{2} =18$

Thus, density at $127^{\circ} C , 18$ option $(c)$ is correct.

$\Rightarrow$  $\frac{{{V_1}}}{{{V_2}}} = \frac{{{T_1}}}{{{T_2}}}$ 

$\Rightarrow$  $\frac{{200}}{{{V_2}}} = \frac{{(273 + 20)}}{{(273 - 20)}} = \frac{{293}}{{253}}$ 

$\Rightarrow$  ${V_2} = \frac{{200 \times 253}}{{293}} = 172.6\,ml$

Let the initial pressure be $P_{1}$ and initial temperature be $T_{1}$.

Thus final pressure $P_{2}=1.004 P_{1}$ and final temperature $T_{2}=T_{1}+1$.

Using $\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$

So, $\frac{ P _{1}}{ T _{1}}=\frac{1.004 P _{1}}{ T _{1}+1}$

Or $T _{1}+1=1.004 T _{1}$

Or $0.004 T _{1}=1$

$\Rightarrow T _{1}=250  \; K$

$\Rightarrow$  $\frac{{{P_1}}}{{{P_2}}} = \frac{{{T_1}}}{{{T_2}}}$

$\Rightarrow$ $\frac{{{P_2} - {P_1}}}{{{P_1}}} = \frac{{{T_2} - {T_1}}}{{{T_2}}}$

$\Rightarrow$  $\left( {\frac{{\Delta P}}{P}} \right)\,\% $$ = \left( {\frac{{251 - 250}}{{250}}} \right) \times 100$ $= 0.4 \%$

Here, at $NTP$, pressure $=76 cm$ of $Hg = P _{1}$

$V _{1}=5 L$

$V _{2}=30 L +5 L$

Now

$P _{1} \times(5)= P _{2} \times(35)$

$P _{1}=7 P _{2}$

$P _{2}=76 / 7=10.8 cm$ of $Hg$

So, pressure in both cylender $=10.8 cm$

$\Rightarrow$  $\frac{{{V_1}}}{{{V_2}}} = \frac{{{T_1}}}{{{T_2}}}$

$\Rightarrow$  $\frac{V}{{{V_2}}} = \frac{{(273 + 27)}}{{(273 + 327)}} = \frac{{300}}{{600}} = \frac{1}{2}$

$\Rightarrow$ ${V_2} = 2V$

$\Rightarrow$  $\frac{{{V_2}}}{{{V_1}}} = \frac{{{T_2}}}{{{T_1}}}$

$\Rightarrow$  $\frac{{{V_2} - {V_1}}}{{{V_1}}} = \frac{{{T_2} - {T_1}}}{{{T_1}}}$

$\Rightarrow$  $\frac{{\Delta V}}{V} = \frac{{(273 + 40) - (273 + 20)}}{{(273 + 20)}}$ 

$ = \frac{{(313 - 293)}}{{293}} = 0.07$

Initial state $P_{0},V_{0},T_{0}\longrightarrow$ Fianl state $1.05 P_{0},V,T_{0}$

Using Boyles law

$P_{1} V_{1}=P_{2} v_{2}$

$P_{0} V_{0}=1.05 P_{0} V$

$V=0.9524 V_{0}$

$\Delta V \%$

$=\frac{V_{f}-V_{i}}{V_{i}} \times 100$

$=\left(\frac{V_{0}-0.9524 V_{0}}{V_{0}}\right) \times 100$

$=4.76 \%$

Temperature of $N _{2}= x$

As $P \propto n T$, as pressure is same.

$\Rightarrow n _{1} T _{1}= n _{2} T _{2} \Longrightarrow\left(\frac{1}{32}\right)(298)=\left(\frac{1}{28}\right)( x )$

$\Longrightarrow x =\frac{28}{32} \times 298=260.75 K$

Temperature in ${ }^{\circ} C =260.75-273=-13^{\circ} C$

At $NTP$,

Pressure, $P=1 \; atm =101325 Pa$

Temperature, $T=20^{\circ} C =\left(20+273.15^{-}\right) K =293.15 K$

Volume given, $V=1$ litre $=10^{-3} m ^{3}$

mass, $m=1.293 gm$

USe equation, $P V=m R T$

$(101325)\left(10^{-3}\right)$

$=(1.293) \cdot R \cdot(293.15)$

$R=0.2673 \; J / K - gm$

$\frac{P_{1}}{\rho_{1} T_{1}}=\frac{P_{2}}{\rho_{2} T_{2}}$

Given that Density at NTP is $\rho_{2}=1.2 \; g / L$

$T _{2}=20^{\circ} C =293.15 K$ and $P _{2}=76 \; cm$ of $Hg$

We have $T _{1}=21^{\circ} C =294.15 \; K$ and $P _{1}=71.8 \; cm$ of $Hg$

Thus $\rho_{1}=\rho_{2} \times \frac{ T _{2}}{ T _{1}} \times \frac{ P _{1}}{ P _{2}}=1.2 \times \frac{293.15}{294.15} \times \frac{71.8}{76}=1.13 \; g / L$

Thus mass of the gas $=\rho_{2} \times V_{2}=1.13 \; g / L \times 1 L =1.13 \; g$

$ = \frac{{(1.64 \times {{10}^{ - 3}} \times 1.01 \times {{10}^5}) \times (1 \times {{10}^{ - 6}})}}{{1.38 \times {{10}^{ - 23}} \times 200}}$

$ = 6.02 \times {10^{16}}$

$\Rightarrow$  $P\left( {\frac{m}{\rho }} \right) = kT$

$\Rightarrow$ $\rho  = \frac{{Pm}}{{kT}}$

$\Rightarrow \frac{{1 \times 500}}{{300}} = \frac{{0.5 \times {V_2}}}{{270}} $

$\Rightarrow {V_2} = 900{m^3}$

$\Rightarrow$ $\frac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \frac{{{\mu _1}}}{{{\mu _2}}}$

$\Rightarrow$ $\frac{{50 \times 100}}{{100 \times {V_2}}} = \frac{1}{2}$

$\Rightarrow$  $V_2$ $= 100\, ml$

$V=\sqrt{\frac{\gamma R T}{M}}$

Volume before expansion $=V_{b}$

Volume after expansion $=V_{a}$

Temp before expansion $=T_{b}$

Temp after expansion $=T_{a}$.

As, Volume os Temperature,

$\frac{V_{b}}{V_{a}}=\frac{T_{b}}{T_{a}}=\frac{V_{b}}{2 V_{b}}=\frac{1}{2}$

$\therefore \frac{T_{b}}{T_{a}}=\frac{1}{2}$

Whow, Before expansion,

$C_{b}=\sqrt{\frac{\gamma R T_{b}}{I M}}$ $\dots \; (1)$

After exparsion,

$C_{a}=\sqrt{\frac{\gamma R T_{a}}{M}}$ $\dots \; (2)$

eq.  $1/2$

$\frac{c_{b}}{c_{a}}=\sqrt{1 / 2} \quad \therefore \quad c_{a}=\sqrt{2} c_{b}$.

$\Rightarrow \frac{{76 \times 1}}{{300}} = \frac{{152 \times 2}}{{{T_2}}}$

$ \Rightarrow {T_2} = 1200K = 927^\circ C$

$m _{1} T _{1}= m _{2} T _{2}$

$13 \times 300= m _{2} \times 325$

$m _{2}=12 \; gm$

Hence mass to be removed $=13-12=1 \; gm$

$\Rightarrow$  $V = \left( {\frac{m}{M}} \right)\frac{{RT}}{P}$=$\left( {\frac{{2.2}}{{44}}} \right) \times \frac{{8.31 \times (273 + 0)}}{{2 \times (1 \times {{10}^5})}}$

= $5.67 \times {10^{ - 4}}{m^3}$ $= 0.56 \; liter$

$\Rightarrow {T_2} = \frac{{{P_2}{V_2}}}{{{P_1}{V_1}}} \times {T_1}$

$ \Rightarrow {T_2} = \frac{1}{{30}} \times \frac{{10}}{1} \times 300 = 100\,K =  - 173^\circ C$

$\Rightarrow {V_2} = \frac{{{P_1}{V_1}}}{{{T_1}}} \times \frac{{{T_2}}}{{{P_2}}}$

$ \Rightarrow {V_2} = \frac{{1500 \times 4 \times 270}}{{300 \times 2}} = 2700{m^3}$

${T_2} = \frac{{{P_2}{V_2}}}{{{P_1}{V_1}}}{T_1} = \frac{2}{1} \times \frac{3}{1} \times 300 $

         $= 1800K = 1527^o C$

$ = m\left( {\frac{R}{M}} \right)T$ $ = \frac{8}{{32}}RT$       $[\because M_{O_2} = 32]$ 

$\Rightarrow$  $PV = \frac{{RT}}{4}$

$Density =( mass ) /( Volume )$

i.e. $\rho=( M / V )$

$\therefore P \times( M / \rho)= nRT$

$\therefore P =\{( nRT \rho) / M \}$

$( P / \rho T )=\text { constant }$

Given: pressure is doubled, Temper attire is $4$ times.

As $\left(P_{2} / P_{1}\right)=\left\{\left(\rho_{2} T_{2}\right) /\left(\rho_{1} T_{1}\right)\right\}$

$\therefore\left\{2 P _{1} / P _{1}\right\}=\left(\rho_{2} / \rho_{1}\right) \cdot\left\{4 T _{1} / T _{1}\right\}$

$\therefore\left(\rho_{1} / \rho_{1}\right)=(1 / 2)$

$\therefore \rho_{2}=\left(\rho_{1} / 2\right)$

Hence density is halved.

$\Rightarrow$  $\frac{{{V_1}}}{{{V_2}}} = \frac{{{T_1}}}{{{T_2}}}$ 

$\Rightarrow$ $\frac{V}{{2V}} = \frac{{(273 + 0)}}{{{T_2}}}$

$\Rightarrow$  ${T_2} = 546\,K = 273^\circ C$

$P' = \frac{{({\mu _1} + {\mu _2})RT}}{V} = \frac{{2PV}}{{RT}} \times \frac{{RT}}{V} = 2P$

$\Rightarrow$ $\frac{{{P_1}}}{{{P_2}}} = \frac{{{m_1}}}{{{m_2}}} \times \frac{{{T_1}}}{{{T_2}}}$

$\Rightarrow$  $\frac{{10}}{{{P_2}}} = \frac{m}{{m/2}} \times \frac{{(273 + 27)}}{{(273 + 87)}}$ 

$\Rightarrow$  ${P_2} = 6\,atm$

$\Rightarrow \frac{{{P_1}}}{{{P_2}}} = \frac{{{T_1}}}{{{T_2}}}$

$\Rightarrow \frac{{{P_1}}}{{(1.01){P_1}}} = \frac{{{T_1}}}{{{T_1} + 5}}$

$ \Rightarrow {T_1} = 500\;K$$= 227°C$

$\Rightarrow$ $P \propto mT$

$\Rightarrow$ $\frac{{{P_2}}}{{{P_1}}} = \frac{{{m_2}}}{{{m_1}}}\frac{{{T_2}}}{{{T_1}}}$$ = \frac{1}{2} \times \frac{{(273 + 27 + 50)}}{{(273 + 27)}} = \frac{7}{{12}}$

$\Rightarrow$  ${P_2} = \frac{7}{{12}}{P_1} = \frac{7}{{12}} \times 20 = 11.67atm. \approx 11.7\,atm$

$\Rightarrow$ $\frac{m}{{VP}}$ $\Rightarrow$  $\frac{{density}}{{P}} = \frac{M}{{RT}}$

${\left( {\frac{{density}}{P}} \right)_{at\,0^\circ C\;}} = \frac{M}{{R(273)}} = x$                 …...$(i)$

${\left( {\frac{{density}}{P}} \right)_{at\,100^\circ C\;}} = \frac{M}{{R(373)}}$                     …..$(ii)$

$\Rightarrow$  ${\left( {\frac{{density}}{P}} \right)_{at\,100^\circ C\;}}$$ = \frac{{273x}}{{373}}$

$ = \frac{{2 \times {{10}^{ - 3}} \times 8.3 \times 300}}{{32 \times {{10}^{ - 3}} \times {{10}^5}}}$=$1.53 \times {10^{ - 3}}{m^3}$ $= 1.53 \,litre$

$n =\frac{ P }{ kT }=\frac{h dg }{ kT }$

Number of molecules in the electron tube is $= n \times V =\frac{ hdgV }{ kT }$

$\Rightarrow n =\frac{ hdg }{ kT }=\frac{1.2 \times 10^{-10} \times 13600 \times 9.8 \times 100 \times 10^{-6}}{1.38 \times 10^{-23} \times 300}=3.86 \times 10^{11}$

$\Rightarrow$  $\frac{P}{{P + \frac{{0.5}}{{100}}P}} = \frac{T}{{T + 2}}$

$\Rightarrow$ $\frac{{200}}{{201}} = \frac{T}{{T + 2}}$

$\Rightarrow$  $T = 400 \,K = 127°C$

$\Rightarrow 1 \times V=n R(35+273)$

$=n R \times 308$        $\dots \; (i)$

$\text { and } \rho_{2} V=n R T_{2}$

$\Rightarrow 3 \times V=n R T_{2}$        $\dots \; (ii)$

$\text { On dividing Eq. (ii) by Eq. (i), we get }$

$3=T_{2} / 308$

$\Rightarrow T_{2}=924 K$

$T_{2}=924-273$

$=651^{\circ} C$

Total pressure of moist gas $=735 \; mm$ of mercury

Aqueous vapour pressure $=23.8 \; mm$

According to Dalton's law of partial pressure,

Total pressure of moist gas $=$ Pressure of dry gas + Aqueous vapour presere

$735 \; mm =$ Pressure of dry gas $+23.8 \; mm$

$\therefore$ Pressure of ary gas $=(735-23.8) \; mm$

Pressure of drygas $=711.2 \; mm$

correct choice - option $-D$

$P_{1}=0.60 \quad \; P_{2}=0.80$

we know that,

$P_{1} v_{1}+P_{2} v_{2} =P\left(v_{1}+v_{2}\right)$

$0.60 \times 125+0.80 \times 150=P(125+150)$

$0.60 \times 0.125 +0.80 \times 0.150=P(0.125+0.150)$

$0.195 =P 00.275$

$P =0.709$

$\Rightarrow$ $\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}$

$\Rightarrow$  $\frac{{200 \times V}}{{(273 + 22)}} = \frac{{{P_2} \times 1.02V}}{{(273 + 42)}}$          $({V_2} = V + 0.02V)$            

$\Rightarrow$ ${P_2} = \frac{{200 \times 315}}{{295 \times 1.02}} = 209\,kPa$

$V _{1}= V , T _{1}=273$

$V _{2}=3 V , T _{2}=?$

We know that at constant pressure

$\frac{ V _{2}}{ T _{2}}=\frac{ V _{1}}{ T _{1}}$

$T _{2}=\frac{3 V \times 273}{ V }$

$T _{2}=819 \; k$

$=546^o$

$\Rightarrow 177^\circ C$

$\Rightarrow \frac{10^{5} \times\left(1 m ^{3}\right)}{300 K }=\frac{ P (2)}{600}$

$\Rightarrow P =10^{5} N / m ^{2}$

Hence,

option $A$ is correct answer.

$(1)$ $1 \; mol =22.4 L$ of gas; $1 \; mol$ of Nitrogen gas $=14 g$

$(2)$ Therefore, $22.4 \; L$ Nitrogen gas $=14 \; g \rightarrow 2 \; L$ Nitrogen $=1.25 \; g$

$(3)$ This is at a standard pressure of $1 \; atm$. So, $22.4 \; atm$ pressure corresponds to $22.4 \times 1.25=28 \; g$

$\Rightarrow$  $\frac{{PV}}{T} = R \approx 2 \frac{{cal}}{{mol{\rm{ - }}K}}$

$V \rightarrow$ volume of the gas

Van der waals equation $\Rightarrow\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$

Coefficient a represents the strength of intermolecular attractive forces. We know that molar volume $\left(V_{m}\right)$ is given by $\Rightarrow V_{m}=\frac{V}{n}$. It shows how the molecules experience the attractive force. The force toward the center experienced by a molecule at the edge should be proportional to $\left(\frac{1}{V_{m}}\right)$. Now, number of molecules at the edge of the container should be proportional to $\left(\frac{1}{V_{m}}\right)$. By combining proportionality constants into a single constant $(a)$

we get,

$\left(\frac{a}{V_{m}^{2}}\right)=\left(\frac{a n^{2}}{V^{2}}\right).$

Answer: Proportion in which press ure will decrease is $\left(\frac{n}{v}\right)^{2}$.

Option $(C)$ is the carrect answer.

$\Rightarrow$ $P = \frac{d}{M}RT$     (density $d = \frac{m}{V}$)

$\Rightarrow$ $\frac{P}{{dT}} = $ constant

$\frac{{{P_1}}}{{{d_1}{T_1}}} = \frac{{{P_2}}}{{{d_2}{T_2}}}$

$PV = RT \Rightarrow V =\left(\frac{ R }{ P }\right) T$

$V \propto T$ (at constant pressure).

Hence, $\frac{ V _{1}}{ V _{2}}=\frac{ T _{1}}{ T _{2}}$

$\Rightarrow \frac{ V _{2}}{ V _{1}}=\frac{ T _{2}}{ T _{1}}$

where, $V _{2}$ is the final volume.

$\frac{ V _{2}}{ V _{1}}-1=\frac{ T _{2}}{ T _{1}}-1$

$\Rightarrow \frac{ V _{2}- V _{1}}{ V _{1}}=\frac{ T _{2}- T _{1}}{ T _{1}} \quad\left[\because T _{2}- T _{1}=1 K \right]$

$\Rightarrow \frac{ V _{2}- V _{1}}{ V _{1}}=\frac{1}{ T _{1}}=\frac{1}{ T }$

$\Rightarrow$  $\frac{{{V_1}}}{{{V_2}}} = \frac{{{m_1}}}{{{m_2}}}.\frac{{{T_1}}}{{{T_2}}}$

$\Rightarrow$ $\frac{{2V}}{V} = \frac{m}{{{m_2}}} \times \frac{{100}}{{200}}$ 

$\Rightarrow$  ${m_2} = \frac{m}{4}$

$\log ( PV )=\log ( nRT )$

$\log P +\log V =\log n+\log R t$

Differentiate it

$\frac{\Delta P }{ P }+\frac{\Delta V }{ V }=\frac{\Delta n }{ n }+\frac{\Delta T }{ T }$

$\because$ It is filled in a rigid cylinder

$\Delta P =0$

$\frac{\Delta V }{ V } \times 100=-10$

$\frac{\Delta T }{ T } \times 100=20$

$\therefore-10=\frac{\Delta n }{ n } \times 100+20$

$\frac{\Delta n }{ n } \times 100=-30$

$\because n =\frac{ m }{ M }$

Where $m \rightarrow$ Mass of gas leaked

$M \rightarrow$ Molar mass

$\therefore \frac{\Delta m }{ m } \times 100=-30$

$\therefore 30 \%$ gas will leak out

$\frac{ P _{1} V _{1}}{ T _{1}}=\frac{ P _{2} V _{2}}{ T _{2}}$

$\frac{1(30000)}{300}=\frac{P(5200)}{143}$

$P =3.86 \; atm$

$T_{A}=T_{B}$

As we know $P V=n R T$

$P_{A} V_{A}=n R T_{A} \; \dots (1)$

$P_{B} V_{B}=n R T_{B} \; \dots (2)$

$T_{A}=T_{B}$

Therefore, $P_{A} V_{A}=P_{B} V_{B}$

Dalton's law of partial pressures that

$p =\left( n _{1}+ n _{2}\right) \frac{ RT }{ V }$

$p _{1}=\frac{ n _{1} RT }{ V }$

$p _{2}=\frac{ n _{1} RT }{ V }$

$n _{1}= n _{2}=\frac{1}{4}$

$p _{1}= p _{2}$

$p _{1}= p _{2}=10$

$p _{1}=5 \; atm$

$T =273 K$ and $P =10^{5} N / m ^{2}( STP )$

From $\quad PV =\mu RT$

$\Rightarrow V =\frac{\mu RT }{ P }=\frac{250 \times 8.3 \times 273}{10^{5}}=5.66 \; m ^{3}$

$P(V - {V_1}) = \frac{m}{M}RT$  $... (i)$

and $P{V_1} = \frac{{2m}}{M}RT$  $... (ii)$

$\Rightarrow$ $\frac{{{V_1}}}{V} = \frac{2}{3}$

$\Rightarrow$ $\tan {\theta _2} > \tan {\theta _1}$ 

$\Rightarrow$ ${\left( {\frac{T}{P}} \right)_2} > {\left( {\frac{T}{P}} \right)_1}$

and $PV = \mu RT$ से $\frac{T}{P} \propto V$ 

$\Rightarrow$ ${V_2} > {V_1}$

Volume $= V$

According to Charle's and Boyle's law,

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

or, $\frac{ PV }{ T }=\frac{2 P \lambda V / 2}{ T _{1}}$

or, $T_{1}=T$

if $T$ is constant,

$PV =$ constant

Hence $\frac{{{P_1}}}{{{P_2}}} = \frac{{{T_1}}}{{{T_2}}}$==> $\frac{1}{3} = \frac{{(273 + 30)}}{{{T_2}}}$

==> ${T_2} = 909K = 636^\circ C$

==> $\frac{{{m_1}}}{{{m_2}}} = \frac{{{P_1}}}{{{P_2}}}$

==> $\frac{{10}}{{{m_2}}}$= $\frac{{{{10}^7}}}{{2.5 \times {{10}^6}}}$==> $m2 = 2.5\, kg$.

Hence mass of the gas taken out of the cylinder

$ = 10 - 2.5 = 7.5kg.$

Volume is also given constant

Hence from $PV = \mu RT = \left( {\frac{m}{M}} \right)RT$==>$T \propto \frac{1}{m}$==> $\frac{{{T_1}}}{{{T_2}}} = \frac{{{m_2}}}{{{m_1}}}$

$\because$ $\frac{1}{4}th$ part escapes, so remaining mass in the vessel ${m_2} = \frac{3}{4}{m_1}$==> $\frac{{(273 + 60)}}{T} = \frac{{3/4\,{m_1}}}{{{m_1}}}$

==> $T = 444K$ $= 171°C$

$T=273 \mathrm{K}$

$P=1.3 \times 10^{5} \mathrm{Pa}$

$\therefore \mathrm{m}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1.3 \times 10^{5} \times 7 \times 10^{-3}}{8.314 \times 273}$

$=0.4$ moles

$\therefore$ No. of molecules $=0.4 \times 6.02 \times 10^{23}$

$=2.4 \times 10^{23}$ molecules

$n_{1}=\frac{35}{28} \times 1000=1250$

$\Rightarrow n_{2}=\frac{1250}{6} \times 9=1250 \times \frac{3}{2}=1875$

$n_{2}=n_{N_{2}}+n_{\circ_{2}}=1875=1250+n_{\mathrm{o}_{2}}$

$n_{\mathrm{o}_{2}}=625$ mole

mass $=$ no. of mole $\times$ molecular weigh $20 \mathrm{kg}$

For $O_{2}$ gas$:$ $P=P, V=V, T=T.$

For $H e$ gas$:$ $P=P_{2}, V=V_{2}=V, T=T_{2}=2 T$

By comparing them $P_{2}=2 P$

For gas $\mathrm{x}: P=P, V=2 V,(n=m / a) ; T=200 \mathrm{K}$

Final position$:$ $P=P, V=V,\left(n=m_{2} / a\right), T=400 K$

By comparing them, $m_{2}=(m / 4)$

$\left(\because \mathrm{v}^{2} \propto \mathrm{T}\right)$

$\frac{P V M}{R_{m}}=T$

$\mathrm{T} \square \mathrm{M}" \mathrm{T}_{\mathrm{He}}<\mathrm{T}_{\mathrm{N}_{2}}$

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

$\frac{1(30000)}{300}=\frac{P(5200)}{143}$

$P=3.86 a t m$

we get $\frac{T_{2}}{T_{1}}=\left(\frac{P_{2}}{P_{1}}\right)\left(\frac{V_{2}}{V_{1}}\right)=\left(\frac{2 P_{1}}{P_{1}}\right)\left(\frac{3 V_{1}}{V_{1}}\right)=6$

$\therefore T_{2}=6 T_{1}=6 \times 300=1800 K=1527^{\circ} \mathrm{C}$

$\frac{\mathrm{V}_{1}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{T}_{2}} \Rightarrow \mathrm{T}_{2}=\frac{\mathrm{V}_{2}}{\mathrm{V}_{1}} \cdot \mathrm{T}_{1}=2(27+273)=600 \mathrm{K}$

$P=$ constant $: V=$ constant

$\mu \mathrm{T}=$ constant $\quad \mathrm{PV}=\mu \mathrm{R} \mathrm{T}$

$\mu_{1} \mathrm{T}_{1}=\mu_{2} \mathrm{T}_{2}$

$\mathrm{x} \times \mathrm{T}=\mu_{2} \times(3 \mathrm{T})$

$\left[\mu_{2}=\frac{\mathrm{x}}{3}\right]$ left in the container

Exit $:$ form container $x-\frac{x}{3}=\frac{2 x}{3}$

$\because P V=R T$

$\frac{R T}{V}=P$

$\because$ For constant $\frac{1}{V} \quad$ So, $P \propto T$

$\because P_C > P_B > P_A$ then

$T_C > T_B > T_A$

After stroke $P V=$ constant

$P V=P_1(V+v)$

$P_1=\frac{P V}{(V+v)}$

Similarly after $2^{\text {nd }}$ stroke

$P_2=P\left(\frac{V}{V+v}\right)^2$

After $n^{\text {th }}$ stroke

$P_n=P\left(\frac{V}{V+V}\right)^n$

Mass $=28 \,g$

$P_i=10 \,atm \quad T_i=57^{\circ} C =330 \,K$

$P_f=5 \,atm \quad T_f=27^{\circ} C =300 \,K$

Volume is kept constant.

$P_i=K \times n_i T_i \quad \dots (i)$

$P_f=K \times n_f T_f \quad \dots (ii)$

Dividing $(i)$ by $(ii)$

$\frac{P_i}{P_f}=\frac{n_i}{n_f} \frac{T_i}{T_f}$

$\frac{n_i}{n_f}=\frac{10}{5} \times \frac{300}{330}$

or $\frac{n_i}{n_f}=2 \times \frac{10}{11}$

$\frac{n_i}{n_f}=\frac{20}{11}$

Now $n_i=1$ mole of $N _2$ $n_f=\frac{11}{20}$ moles

or Mass of $N _2$ left $=\frac{11}{20} \times 28$

$\therefore \text { Quantity released }=28-\frac{11}{20} \times 28$

$=\frac{9}{20} \times 28=\frac{63}{5} g$

$P V=$ constant

$P_1 V_1=P_2 V_2 \quad\left[P_1=P_0\right.$ atmospheric pressure $]$

$P_0 \times 40=P_1 \times 60 \quad \ldots (i)$

$P_1+20=P_0 \quad... (ii)$

From $(i)$

$P_1=\frac{2 P_0}{3}$

From $(ii)$

$\frac{2 P_0}{3}+20=P_0 \Rightarrow P_0=60 cm \text { of } Hg .$

$\frac{P}{T}=$ constant

Initially $\frac{P_0}{T_0}+\frac{P_0}{T_0}=\frac{2 P_0}{T_0}$

For two containers

$\frac{P_0}{T_0}+\frac{P}{T}=\frac{2 P_0}{T_0}$

$P=\frac{2 P_0 \times T \times T_0}{T_0\left(T+T_0\right)}$

or $P=\frac{2 P_0 T}{\left(T+T_0\right)}$

$\Rightarrow$ ${v_{rms}} \propto \sqrt {\frac{P}{m}} $

$\Rightarrow$ $\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{P_1}}}{{{P_2}}} \times \frac{{{m_2}}}{{{m_1}}}} $ $\Rightarrow$ $\frac{v}{{2v}} = \sqrt {\frac{{{P_0}}}{{{P_2}}} \times \frac{{m/2}}{m}} $ $\Rightarrow$  ${P_2} = 2{P_0}$

 $ = \frac{{{{10}^5}}}{{20 \times {{10}^{ - 2}}}} = 5 \times {10^6}N/{m^2}$

$\Rightarrow$ $P = \frac{{v_{rms}^2\rho }}{3} = \frac{{{{(3180)}^2} \times 8.99 \times {{10}^{ - 2}}}}{3}$

$ = 3 \times {10^5}\,N/{m^2} = 3\,atm$

where, $\mu=$ viscerity $P=$ Pressure $R=$ gass constant $T=$ Temperature $M=$Molecular weight

$\Rightarrow \lambda \alpha \frac{1}{p}$

If $p \propto \frac{1}{\lambda}$

If we double the mean force

Path, the pressure will be $1 / 2$.

So, The correct answer is $\frac{p}{2}$

$\Rightarrow$ $PV = \frac{2}{3}E$

 $ = \frac{3}{2}RT = \frac{3}{2} \times 8.3 \times 200 = 2490\,J$

$\Rightarrow$ $\frac{{{E_1}}}{{{E_2}}} = \frac{{{T_1}}}{{{T_2}}} = \frac{{(273 - 23)}}{{(273 + 227)}} = \frac{{250}}{{500}} = \frac{1}{2}$

$\Rightarrow$ ${E_2} = 2{E_1} = 2 \times 5 \times {10^{ - 14}} = 10 \times {10^{ - 14}}\,erg$

$\frac{{{E_1}}}{{{E_2}}} = \frac{{{T_1}}}{{{T_2}}}$ Þ $\frac{E}{{2E}} = \frac{{{T_1}}}{{{T_2}}}$ ${T_2} = 2{T_1}$

${T_2} = 2(273 + 20) = 586\,K = 313^\circ C$

Mass of oxygen, $m=20 gm$

No. of moles, $n=\frac{m}{M}=\frac{m}{\text { molar mass }}$

$\therefore n=\frac{20}{32}$

Temperature, $T=47^{\circ} C =320 K$

The kinetic energy of $0_{2}, E=\frac{3}{2} n R T$

$E=\frac{3}{2} \times\left(\frac{20}{32}\right) \times(8.3) \times(320)$

$E=2490 \; J$

the kintic energy of $0_{2}$ is $2490 \; J$

$R =8.31 J / molK$

$T =273 \;K$

Kinetic energy,

$K =\frac{3}{2} RT$

$K =\frac{3}{2} \times 8.31 \times 273=3.4 \times 10^{3} \; J$

$\Rightarrow$  $\frac{{{E_1}}}{{{E_2}}} = \frac{{{T_1}}}{{{T_2}}}$ $\Rightarrow$  $\frac{{100}}{{{E_2}}} = \frac{{300}}{{450}}$ $\Rightarrow$  ${E_2} = 150\,J$

$\Rightarrow$ $\frac{E}{{2E}} = \frac{{(273 + 27)}}{{{T_2}}}$

$\Rightarrow$ ${T_2} = 600\,K = 327^\circ C$

So, $\frac{K E_{1}}{K E_{2}} =\frac{T_{1}}{T_{2}} \Rightarrow \frac{(27+273) K}{(327+273) K}=\frac{E_{1}}{E_{2}}$

$\frac{E_{1}}{E_{2}} =\frac{300}{600}=\frac {1}{2} \Rightarrow {E_{2}=2 E_{1}}$

So, $K E_{avg}$ for $H_{2}=$ $KE_{avg}$ for He.

and $E_{avg}$ for $H _{2} > E_{avg}$ for $He [\because H _{2}$ has $5$ degree of freedom$]$

$\Rightarrow 2 = \frac{{{T_2}}}{{{T_1}}} $

$\Rightarrow {T_2} = 2{T_1}$ 

$ \Rightarrow {T_2} = 2(273 - 68) = 410K = 137^\circ C$

$\Rightarrow$  $\frac{{{E_1}}}{{{E_2}}} = \frac{{{T_1}}}{{{T_2}}} = \frac{{300}}{{350}} = \frac{6}{7}$

$\Rightarrow {E_2} = {E_1}\frac{{{T_2}}}{{{T_1}}} = 100 \times \frac{{450}}{{300}} = 150J$

$ = \left( {\frac{6}{{32}} + \frac{8}{{28}} + \frac{5}{{44}}} \right) \times \frac{{8.31 \times 300}}{{3 \times {{10}^{ - 3}}}} = 5 \times {10^5}N/{m^2}$

$\vec{P}_{a v}=M\vec V_{a v}$

$\left[\vec{V}_{a v}=0\right. Average\, velocity\, of\, gas\, molecules\, at\, any\, temperature\, is\, always\, zero]$

Hence, $\vec{P}_{a v}=0$

 $\frac{f}{2}({n_1} + {n_2})kT$$ = \frac{f}{2}{n_1}k{T_1} + \frac{f}{2}{n_2}k{T_2}$ 

$\Rightarrow$  $T = \frac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}}$

Where, $f$ is the degree of frecdem.

So, $\frac{K E _m}{f}=\frac{K_ B T}{2}$

$\Rightarrow$ $T = \frac{2}{3}\frac{{eV}}{k} = \frac{{\frac{2}{3} \times 1.6 \times {{10}^{ - 19}}}}{{1.38 \times {{10}^{ - 23}}}} = 7.7 \times {10^3}K$

$ = \frac{{8.99 \times {{10}^{ - 2}} \times 1840 \times 1840}}{3}$$ = 1.01 \times {10^5}N/{m^2}$

$\Rightarrow$ $\frac{{E'}}{E} = \frac{{T'}}{T} = \frac{{400}}{{300}} = \frac{4}{3} = 1.33$

$\therefore C _{ V }= M _{ CV } 4 \times 3=12 \; J mol ^{-1} K ^{-1}$

At constant volume $P \propto T$.

$\therefore \frac{ P _{2}}{ P _{1}}=\frac{ T _{2}}{ T _{1}}=2, T _{2}=2 T _{1}$

Rise in temperature $\Delta T = T _{2}- T _{1}=2 T _{1}- T _{1}= T _{1}=273 \; K$

Heat required, $\Delta Q = NC V \Delta T =\frac{1}{2} \times 12 \times 273=1638 \; J$

$\Rightarrow$ $\frac{{{T_2}}}{{{T_1}}} = \frac{{{N_1}}}{{{N_2}}} = \frac{1}{2}$ $\Rightarrow$ ${T_2} = \frac{{{T_1}}}{2}$

and $PV =  NkT$ $\Rightarrow$ $P \propto NT$

$\Rightarrow$ $\frac{{{P_1}}}{{{P_2}}} = \frac{{{N_1}}}{{{N_2}}}.\frac{{{T_1}}}{{{T_2}}} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$

$\therefore E=\mu \frac{7}{2} R T=\left(\frac{m}{M}\right) \frac{7}{2} N k T$

$=\frac{1}{44}\left(\frac{7}{2}\right) N k T=\frac{7}{88} N k T$

[As $f=7$ and $M=44$ for $CO _{2}$ ]

$Q = nC _{ p } \Delta T$

$40=1 \times C_{p} \times 10$

Thus, $C _{ p }=4$

Now, $\gamma=\frac{ C _{ p }}{ C _{ v }}$

$\gamma$ for monoatomic gas $=1.67$

Thus, $C _{v}=\frac{ C _{p}}{\gamma}=\frac{4}{1.67}=2.4$

For a constant volume process,

$Q = nC _{ v } \Delta T$

Thus,

$Q=1 \times 2.4 \times 10$

$Q =24 cal \approx 20 cal$

$\Rightarrow$ $W = \frac{{{{(\Delta Q)}_V}}}{{\Delta T}} = \frac{{\Delta U}}{{\Delta T}}$     $[\because \,\,{(\Delta Q)_V} = \Delta V]$

$\Rightarrow$  $W = \frac{{80}}{{(120 - 100)}} = 4\,J/K$

No. of moles $= n =5$,

change in temperature $=\Delta T =10 K$

Amount of heat absorbed $=\Delta Q = nC _{v} \Delta T$

$=(5 \times 5 \times 10) \; cal$

$=250 \; cal$

where $C$ is the mean kinetic energy of molecule at 0 $K$ temperature.

So, $EVs T$ is a straight line having intercept on positive $y$-axis as shown in figure $C$.

To Find :- Value of $a$ and $b$ in graph

Solution :- Since we know that Volume cannot be negative any more

$\therefore$ Here, $(b)$ is Volume

($\because$ a also contains negative value)

Since we know that Kelvin Temperature cannot be negative

$\therefore$ a is temperature $\left({ }^{\circ} C \right)$

$\frac{ PV }{ nT }= R \quad$ (R here is a constant)

$ \frac{ P _{1} V _{1}}{ n _{1} T _{1}}=\frac{ P _{2} V _{2}}{ n _{2} T _{2}}$

$P _{1}=760 mm$ of $Hg$

$V _{1}=22400 cm ^{3}$

$T _{1}=273 K$

$P _{2}=10^{-11} mm$ of $Hg$

$T _{2}=300 K$

$n _{1}=6.02 \times 10^{23}$

$\frac{ n _{2}}{ V _{2}}=\frac{ P _{2} n _{1} T _{1}}{ P _{1} V _{1} T _{2}}$

$\frac{ n _{2}}{ V _{2}}=\frac{10^{-11} \times 6.02 \times 10^{23} \times 273}{760 \times 22400 \times 300}=3.32 \times 10^{5} \;molecules / cm ^{3}$

$\mathrm{n}=\frac{2 \times 10^{5} \times 8 \times 10^{-3}}{8.31 \times 300}=0.64 \mathrm{\,mole}$

at same Temp. $P \propto \mu$

$\frac{P_{2}}{P_{1}}=\frac{\mu_{2}}{\mu_{1}} $

$\Rightarrow \frac{125 \times 10^{3}}{2 \times 10^{5}}=\frac{\mu_{2}}{0.64} $

$\Rightarrow \mu_{2}=0.4$ $mole$

leaked moles $=0.64-0.40=0.24$ $moles$

$P \propto mT \Rightarrow \frac{{{P_2}}}{{{P_1}}} = \frac{{{m_2}}}{{{m_1}}}\frac{{{T_2}}}{{{T_1}}} = \frac{3}{4} \times \frac{{626}}{{313}}= 1.5$

==> ${P_2} = 1.5{P_1} = 1.5 \times 720 = 1080\,kPa$

Hence number of moles of gas $\mu = \frac{{44.8}}{{22.4}} = 2$

Since the volume of cylinder is fixed,

Hence ${(\Delta Q)_V} = \mu \omega \;\Delta T$

$ = 2 \times \frac{3}{2}R \times 10 = 30R$

==> $\frac{{{{(\Delta Q)}_V}}}{{{{(\Delta Q)}_P}}} = \frac{{{C_V}}}{{{C_P}}} = \frac{{\frac{3}{2}R}}{{\frac{5}{2}R}} = \frac{3}{5}$ $\left[ {\because \,{{({C_V})}_{mono}} = \frac{3}{2}R,\,{{({C_P})}_{mono}} = \frac{5}{2}R} \right]$

==> ${(\Delta Q)_V} = \frac{3}{5} \times {(\Delta Q)_P} = \frac{3}{5} \times 210 = 126\,J$

$ = 2$ litre=$2 \times {10^{ - 3}}$${m^3}$

So pressure at top $=$ atmospheric pressure $-50 \mathrm{cm}$ of $\mathrm{Hg}=(75-50) \mathrm{cm}=25 \mathrm{cm}$ of $\mathrm{Hg}=$ $1 / 3 \times 100 k P a=33.33 k P a$

Hence, at constant pressure, $\frac{V}{T}=\frac{m}{M} \frac{R}{P}$

In the second case,

$\frac{V}{T}=\frac{2 m}{M} \frac{R}{2 P}=\frac{m}{M} \frac{R}{P}$

since the slope in two cases is same. Hence, new line will also be $B$. Curve does not pass through $0,0$ due to non idealities at low temperature and pressure.

Answer is option $C.$

Translational degree of freedoms $=3$

Rotational degree of freedoms $=2$

Ratio $=\frac{3 \mathrm{KT}}{2 \mathrm{KT}}=\frac{3}{2}$

$\frac{1}{2} m v^{2}=n C_{v} \Delta T$

$\frac{1}{2} \times \frac{30}{1000} \times 100^{2}=1 \times \frac{R}{\gamma-1} \times \Delta T$

Here $\gamma$ is 1.4 for a diatomic gas, so $150=\frac{R}{1.4-1} \times \Delta T$

$\Delta T=\frac{60}{R}$

Internal energy $\quad H=200 J$

while mixing, they don't interact Internal energy of $\operatorname{mix}(100+200) J=300 J$

As the process is $\frac{P}{V}=$ constant

i.e., $P V^{-1}=$ constant, therefore, $x=-1$

For an ideal monatomic gas, $\gamma=\frac{5}{3}$

$\therefore C=\frac{R}{\frac{5}{3}-1}+\frac{R}{1-(-1)}=\frac{3}{2} R+\frac{R}{2}=2 R$

$\Delta Q=n C(\Delta T)=1(2 R)\left(2 T_{0}-T_{0}\right)=2 R T_{0}$

$K E=\frac{3}{2} R T$ so

$K E \propto T$

means, mean transitional kinetic energy of the molecules is proportional to the absolute temperature.

Thus amount of gas escaped is $n-0.8 n=0.2 n$ or $20 \%$

So, the kinetic energy is proportional to the temperature.

The per unit mass of molecule will change So, the volume and momentum will also be changed.

Thus, in case of hydrogen and oxygen at NTP the average kinetic energy per molecule will be same.

$\Rightarrow \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\rho_{1} \mathrm{M}_{2}}{\rho_{2} \mathrm{M}_{1}}=\frac{4}{3} \Rightarrow \frac{\rho_{1}}{\rho_{2}}=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9}$

$=\left(\frac{3}{2}+1\right)(2 \times 8.31)(5)=208 \mathrm{J}$

But $P V=\mu R T$

$V=\frac{\text { mass }}{\text { density }}=\frac{1 \mathrm{kg}}{4 \mathrm{kg} / \mathrm{m}^{3}}=\frac{1}{4} \mathrm{m}^{3}$

$P=8 \times 10^{4} \mathrm{N} / \mathrm{m}^{2}$

$\therefore U=\frac{5}{2} \times 8 \times 10^{4} \times \frac{1}{4}=5 \times 10^{4} \mathrm{J}$

$\mathrm{K}_{\mathrm{av}}=\frac{3}{2} \mathrm{kT}\{\mathrm{Ar}\}$

$\mathrm{p}_{1}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}} \times \mathrm{p}$

$\mathrm{n}_{1}>\mathrm{n}_{2} \Rightarrow \mathrm{p}_{1}>\mathrm{p}_{2}$

$\mathrm{Q}=\mathrm{RT}$

$\Delta \mathrm{KE}=\Delta \mathrm{U}$

$\Rightarrow \frac{1}{2} \mathrm{mV}^{2}=\mu \mathrm{C}_{\mathrm{v}} \mathrm{d} \mathrm{T}\left[\mathrm{C}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}\right]$

$\Rightarrow \frac{1}{2} \mathrm{mV}^{2}=\frac{\mathrm{m}}{\mathrm{M}} \times \frac{5}{2} \mathrm{R} \times \Delta \mathrm{T}$

$\Rightarrow \Delta \mathrm{T}=\frac{\mathrm{MV}^{2}}{5 \mathrm{R}}$

$\frac{\mathrm{U}_{\mathrm{O}_{2}}}{\mathrm{U}_{\mathrm{He}}}=\frac{\left[\frac{(1)(5) \mathrm{RT}}{2}\right]}{\left[\frac{(2)(3) \mathrm{RT}}{2}\right]}=\frac{5}{6}$

$\Delta \mathrm{U}=\frac{\mathrm{f}}{2}\left(\mathrm{P}_{2} \mathrm{V}_{2}-\mathrm{P}_{1} \mathrm{V}_{1}\right)=\frac{5}{2}(4 \times 5-8 \times 2) \times 10^{3}$

$\Delta \mathrm{U}=\frac{5}{2}(20-16) \times 10^{3}=10 \mathrm{kJ}$

explanation$:$ when a gas has volume $\mathrm{v}$ and pressure $\mathrm{P}$, total translational kinetic energy of all molecules of gas is $3 / 2 P V$.

actually, Translational kinetic energy depends upon translational degree of freedom. and we know, there are only three degree of freedom in translational motion for any gas (monoatomic, diatomic or triatomic etc.)

Hence under same pressure, equal volumes of gaseous molecules have same translational kinetic energy i.e. $(3 / 2) P V$

$E=\frac{3}{2} \mathbf{P V}$

where $E=$ total translational $\mathrm{K} E$

for Both gas slope will be same.

$\therefore \Delta \mathrm{T}=\left(\frac{25}{\mathrm{nR}}\right)$

Now, for above process $f=6$

So, $C_{P}\left(1+\frac{f}{2}\right) R=4 R$

Heat absorbed $=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}=\mathrm{n} \times 4 \mathrm{R} \times \frac{25}{\mathrm{nR}}=100 \mathrm{J}$

$\Delta Q=n C_P \Delta T$

$20=n C_P \times 10 \quad \ldots(1)$

$\Delta U=n C_V \Delta T$

$\Delta U=n \frac{C_P}{\gamma} \Delta T$ $\left\{\because \gamma_{\text {mono }}=5 / 3\right\}$

$\Delta U=12 \,J$

Heat supplied $=\Delta U=U_{\text {final }}-U_{\text {initial }}$

$U_{\text {initial }}=\frac{5}{2} \times 20 \times R T, U_{\text {final }}=\frac{5}{2} \times(20-8) R T+\frac{3}{2} \times(2 \times 8) R T$

$\Delta U=\frac{1}{2} \times 8 \times R T$

$=4 R T$

$\Rightarrow$ Heat energy given is $4 R T$.

$\frac{\Delta U}{\Delta Q}=\frac{n C_V \Delta T}{n C_P \Delta T}=\frac{C_V}{C_P}=\frac{1}{\gamma}$

$\left(\frac{\Delta U}{\Delta Q}\right)_{\text {mono }}=\frac{1}{\gamma_{\text {mono }}}=\frac{3}{5}$

$\left(\frac{\Delta U}{\Delta Q}\right)_{\text {dia }}=\frac{1}{\gamma_{\text {dia }}}=\frac{5}{7}$

$\left(\frac{\Delta U}{\Delta Q}\right)_{\text {tria }}=\frac{1}{\gamma_{\text {tria }}}=\frac{3}{4}$

Fractional energy used to change internal energy is maximum in Triatomic gas.

At constant pressure heat absorbed $=\Delta Q=n C_P \Delta T \quad \dots (1)$

At constant volume heat absorbed $=\Delta U=n C_v \Delta T \quad \dots (2)$

Dividing $(1)$ by $(2)$,

$\frac{\Delta Q}{\Delta U}=\frac{C_P}{C_V}=\gamma=1.4 \Rightarrow \frac{105}{\Delta U}=1.4$

$\therefore \Delta U_V=75 \,cal$

Heat at constant pressure

$\Delta Q=n C_p \Delta T$

Heat for doing work

$\Delta W=n R \Delta T$

Then $\frac{\Delta W}{\Delta Q}=\frac{n R \Delta T}{n C_P \Delta T}$

$\frac{\Delta W}{800}=\left(\frac{\gamma-1}{\gamma}\right)$

$\frac{\Delta W}{800}=1-\frac{1}{\gamma}$

$\frac{\Delta W}{800}=1-\frac{3}{4}$

$\Delta W=200 \,cal$

Heat supplied $=\Delta U=U_{\text {final }}-U_{\text {initial }}$

Total internal energy initially $=\frac{5}{2} N R T$  [Only diatomic gas is present]

Total internal energy when

' $n$ ' moles get dissociated $=\frac{5}{2}(N-n) R T+\frac{3}{2}(2 n) R T$ [diatomic and monoatomic both are present]

$\Delta U=\left\{\frac{5}{2}(N-n) R T+\frac{3}{2}(2 n) R T\right\}-\frac{5}{2} N R T$

Solving this we get

$\Delta U=\frac{1}{2} n R T$

$\therefore$ Heat supplied is $\frac{1}{2} n R T$.

Initial energy of gas = Final energy

Let K.E. of each molecule initially be $E_0$

$\therefore$ Total kinetic energy $=E_0 \times n$

Let final kinetic energy of each molecule be $E_f$

$E_0 \times n=E_f \times 2 n$

$E_f=\frac{E_0}{2}$

Since temperature is the average kinetic energy of molecules.

$T_0=K E_0 \quad T_f=\frac{K E_0}{2}$

$\therefore$ Temperature becomes $T_f=\frac{T_0}{2}$

Degrees of freedom of $He \left(f_{ He }\right)=3$

Degrees of freedom of $O _2\left(f_{ O _2}\right)=5$

Degrees of freedom of $O _3\left( f _{ O _3}\right)=6$

$n_{ He }=2, \quad n_{ O _2}=4 \quad n_{ O _3}=1$

Energy of mixture = Sum of individual energies

$=\left(n_{ He}f_{ He }+n_{ O _2} f_{ O _2}+n_{ O _3} f_{ O _3}\right) \frac{R T}{2}$

$=(2 \times 3+4 \times 5+1 \times 6) \frac{R T}{2}$

$=(3+10+3) R T$

$=16 R T$

$1$ mole of $CO$ and $1$ mole of $O _2$ are mixed.

Net internal energy $=\frac{f_1}{2} R T_{ CO }+\frac{f_2}{2} R T_{ O _2}$

$=\frac{5}{2} R 300+\frac{5}{2} R 350$

$=\frac{5}{2} R(650)$

$=5 R(325)$

$=1625 R$

$1625=\frac{5}{2} R T_{\text {final }} \times n_{\text {final }}$

$\frac{1625 \times 2}{5}=T_{\text {final }} \times n_{\text {final }}$

$325 \times 2=T_{\text {final }} \times 2$

$T_{\text {final }}=325 K$

$T_{\text {final }}=37^{\circ} C$

Variation of atmospheric pressure due to height is given by the barometric formula

$P_h=P_0 e ^{-m g h / R T}$

Hence the decrease will be exponential.

Let there by $n$ moles of gas.

Mass of gas $=40 n g$ or $\frac{40 n}{1000}$ or $0.04 n kg$

$\text { K.E. of gas in container }=\frac{1}{2} \times 0.04 n \times(200)^2$

$=0.02 \times n \times 4 \times 10^4$

$=8 \times 10^2 \times n J$

Now heat capacity of gas $(C)=\frac{f}{2} n R$

or $C=\frac{5}{2} R \times n$

or $C \Delta T=8 \times 10^2 \times n$

or $\frac{5}{2} \times R \times n \Delta T=8 \times 10^2 \times n$

$\Delta T=\frac{8 \times 10^2}{R} \times \frac{2}{5}$

$\Delta T=\frac{16}{5} \times 10^2=\frac{320}{R}{ }^{\circ} C$

We know that

$P_A V_A=n_A R T, P_B V_B=n_B R T$

and $P_f\left(V_A+V_B\right)=\left(n_A+n_B\right) R T$

$P_f\left(V_A+V_B\right)=P_A V_A+P_B V_B$

$\therefore P _{ f } =\left(\frac{ P _{ A } V _{ A }+ P _{ B } V _{ B }}{ V _{ A }+ V _{ B }}\right)$

$=\frac{1.4 \times 0.1+0.7 \times 0.15}{0.1+0.15} MPa =0.98\,MPa$

Energy possessed by the ideal gas at $27^{\circ}\,C$ is

$E _1=3\left(\frac{3}{2} R \times 300\right)=\frac{2700 R }{2}$

Energy possessed by the ideal gas at $227^{\circ} C$ is

$E _2=2\left(\frac{3 R }{2} \times 500\right)=1500\,R$

If $T$ be the equilibrium temperature, of the mixture, then its energy will be $E _{ m }=5\left(\frac{3 RT }{2}\right)$

Since, energy remains conserved,

$E _{ m }= E _1+ E _2$

or $5\left(\frac{3 R T}{2}\right)=\frac{2700 R }{2}+1500 R$ or $T =380 K$ or $107^{\circ}\,C$

$\Rightarrow$  $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^2} = \frac{1}{4}$

$\Rightarrow$  ${T_2} = $$\frac{{{T_1}}}{4} = \frac{{(273 + 327)}}{4} = 150K =  - {123^o}C$

$ \Rightarrow \,\,\frac{{{{({v_{rms}})}_{{O_2}}}}}{{{{({v_{rms}})}_{{H_2}}}}} = \sqrt {\frac{{{M_{{H_2}}}}}{{M{o_2}}}} \, \Rightarrow \,\frac{{400}}{{{{({v_{rms}})}_{{H_2}}}}} = \sqrt {\frac{2}{{32}}}  = \frac{1}{4}$

$ \Rightarrow \,\,\,{({v_{rms}})_{{H_2}}} = 1600\,\,m/s$

$ \Rightarrow \,\,\,\frac{{{T_{{O_2}}}}}{{{T_{{H_2}}}}} = \frac{{{M_{{O_2}}}}}{{{M_{{H_2}}}}} \Rightarrow \,\,\frac{{{T_{{O_2}}}}}{{200}} = \frac{{32}}{2}\, \Rightarrow \,\,{T_{{O_2}}} = 3200\,K$

$\frac{{{v_{He}}}}{{{v_{{H_2}}}}} = \frac{5}{7} = \sqrt {\frac{{{T_{He}}}}{{{M_{He}}}} \times \frac{{{M_{{H_2}}}}}{{{T_{{H_2}}}}}} $

$\Rightarrow$  ${T_{He}} = \frac{{25}}{{49}} \times \frac{4}{2} \times 273$

$= 273K$ = $0^\circ C$

$T_{2}=?$

That the Final $rms$ speed will be Double of inital.

$\frac{V_{rms_{2}}}{V_{rms_{1}}}=\sqrt{\frac{T_{2}}{T_{1}}}=\sqrt{\frac{T_{2}}{273}}$

$V_{rms_{2}} =2 V_{rms_{1}}$

$\Rightarrow \quad 2 =\sqrt{\frac{T_{2}}{273}}$

$\Rightarrow \quad T_{2} =4 \times 273=1092 \; K.$

$\quad T_{2} =819^{\circ} C$

The temperature raised $=819^{\circ} C$

How, when other terms are constant, $(R \cdot M . S) \propto \sqrt{T}$.

$\frac{(R \cdot M \cdot S)_{1}}{(R \cdot M \cdot S)_{2}}=\frac{\sqrt{T_{1}}}{\sqrt{T_{2}}}$

Here, $(R \cdot M \cdot S)_{2}=2 \cdot(R \cdot M \cdot S)_{1}$

$T_{1} =100^{\circ} C =(100+273) K =373 \; K$

$\therefore \frac{(R \cdot M \cdot S)_{1}}{2 \cdot(R \cdot M \cdot S)} =\frac{\sqrt{373}}{\sqrt{T_{2}}}$

$\sqrt{T_{2}} =2 \cdot \sqrt{373}$

$T_{2} =1492 \; K$ [squaring both sides]

Convert into celcius.,

$T_{2}=(1492-273)^{\circ} C =1219^{\circ} C$

$\therefore$ $\frac{{{v_{rms}}}}{{{v_s}}} = \sqrt {\frac{\gamma }{3}} $

$\Rightarrow$ $\frac{{{T_{{O_2}}}}}{{{T_{{H_2}}}}} = \frac{{{M_{{O_2}}}}}{{{M_{{H_2}}}}}$

$\Rightarrow$ $\frac{{{T_{{O_2}}}}}{{(273 + 0)}} = \frac{{32}}{{28}}$

$\Rightarrow$ ${T_{{O_2}}} = 312\,K \to {39^o}C$

${v_{av}} = \frac{{1 + 2 + 3 + 4 + 5}}{5} = 3km/s$

$\Rightarrow$ $\frac{{{v_{rms}}}}{{{v_{av}}}} = \frac{{\sqrt {11} }}{3}$

$\Rightarrow$  $\frac{{{v_{{H_2}}}}}{{{v_{He}}}} = \sqrt {\frac{{{M_{He}}}}{{{M_{{H_2}}}}}}  = \sqrt {\frac{4}{2}}  = \sqrt 2 $

 ${v_{{H_2}}} = \sqrt 2 \,{v_{He}}$

$\frac{{{v_2}}}{{{v_1}}} = \sqrt {\frac{{{T_2}}}{{{T_1}}}} $

$\Rightarrow$ ${v_2} = \sqrt {\frac{{(273 + 927)}}{{(273 + 2)}}} $

$\Rightarrow$ ${v_2} = 2{v_1}$

$\therefore$  $\frac{{{{({v_{rms}})}_{{O_2}}}}}{{{{({v_{rms}})}_{{H_2}}}}} = \sqrt {\frac{{{M_{{H_2}}}}}{{{M_{{O_2}}}}}}  = \sqrt {\frac{2}{{32}}}  = 1:4$

$\Rightarrow$ $T \propto M$               

$\Rightarrow$ $\frac{{{T_{{H_2}}}}}{{{T_{{O_2}}}}} = \frac{{{M_{{H_2}}}}}{{{M_{{O_2}}}}}$Þ$\frac{{{T_{{H_2}}}}}{{(273 + 31)}} = \frac{2}{{32}}$

$\Rightarrow$ ${T_{{H_2}}} = 19\,K =  - \,254^\circ C$

$\Rightarrow$ $T \propto M$

$\Rightarrow$ $\frac{{{T_{He}}}}{{{T_H}}} = \frac{{{M_H}}}{{{M_{He}}}}$

$\Rightarrow$ $\frac{{(273 + 0)}}{{{T_H}}} = \frac{2}{4}$

$\Rightarrow$ ${T_H} = 546K = 273^\circ C$

average velocity of a gas molecule

 ${v_{av}} = \sqrt {\frac{{8RT}}{{\pi M}}} $

$\Rightarrow$ ${v_{av}} \propto \frac{1}{{\sqrt M }}$

$\Rightarrow$ $\frac{{ < {C_H} > }}{{ < {C_{He}} > }} = \sqrt {\frac{{{M_{He}}}}{{{M_H}}}}  = \sqrt {\frac{4}{1}}  = 2$ 

$\Rightarrow$  $ < {C_H} >  = 2\, < {C_{He}} > $

 $ \Rightarrow$ $\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} $$ = \sqrt {\frac{{(273 + 27)}}{{(273 + 327)}}}  = \sqrt {\frac{{300}}{{600}}}  = \frac{1}{{\sqrt 2 }}$

 $ \Rightarrow \,\,{v_2} = \sqrt 2 {v_1}$