3 Marks Question

Question 13 Marks

A vessel containing one mole of a monatomic ideal gas (molecular weight $=20 \mathrm{gmol}^{-1}$ ) is moving on a floor at a speed of $50 \mathrm{~ms}^{-1}$. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

Answer

N = 1mole, M = 20g/mol, V = 50m/s K.E. of the vessel = Internal energy of the gas$=\Big(\frac{1}{2}\Big)\text{mv}^2$
$=\Big(\frac{1}{2}\Big)\times20\times10^{-3}\times50\times50=25\text{J}$
$\Rightarrow25=\text{n}\frac{3}{2}\text{r}(\triangle\text{T})$
$\Rightarrow25=1\times\frac{3}{2}\times8.31\times\triangle\text{T}$
$\Rightarrow\triangle\text{T}=\frac{50}{3\times8.3}\approx2\text{k}$

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MCQ 23 Marks

An ideal gas expands from $100\ cm^3$ to $200\ cm^3$ at a constant pressure of $2.0 \times 10^5 Pa$ when $50J$ of heat is supplied to it. Calculate,

A
The change in internal energy of the gas.
B
The number of moles in the gas if the initial temperature is $300K.$
C
The molar heat capacity $C_p$ at constant pressure.
✓
The molar heat capacity $C_v$ at constant volume.

Answer

Correct option: D.

The molar heat capacity $C_v$ at constant volume.

$V_1 = 100\ cm^3, V_2= 200\ cm^3, P = 2 \times 10^5Pa,$

$\triangle\text{Q}=50\text{J}$

$\triangle\text{Q}=\text{du}+\text{dw}$
$\Rightarrow50=\text{du}+20\times10^5(200-100\times10^{-6})$
$\Rightarrow50=\text{du}+20$
$\Rightarrow\text{du}=30\text{J}$

$30=\text{n}\times\frac{3}{2}\times8.3\times300$ $\Big[\text{U}=\frac{3}{2} \text{nRT}$ for monoatomic$\Big]$

$\Rightarrow\text{n}=\frac{2}{3\times83}=\frac{2}{249}=0.008$

$\text{du}=\text{nC}_\text{V}\text{dT}$

$\Rightarrow\text{C}_\text{V}=\frac{\text{dU}}{\text{ndT}}$
$=\frac{30}{0.008\times300}=12.5$
$\text{C}_\text{P}=\text{C}_\text{V}+\text{R}=12.5+8.3=20.8\text{J/mol-K}$

$\text{C}_\text{V}=12.5\text{J/mol-K}$

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Question 33 Marks

An ideal gas $\Big(\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=\gamma\Big)$ is taken through a process in which the pressure and the volume vary as $p = aV^b$. Find the value of b for which the specific heat capacity in the process is zero.

Answer

$\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=\gamma,\ \text{C}_\text{p}-\text{C}_\text{V}=\text{R},$$\text{C}_\text{V}=\frac{\text{r}}{\gamma-1},\text{C}_\text{P}=\frac{\gamma\text{R}}{\gamma-1}$
$\text{Pdv}=\frac{1}{\text{b}+1}(\text{Rdt})$
$\Rightarrow0=\text{C}_\text{V}\text{dT}+\frac{1}{\text{b}+1}(\text{Rdt})$
$\Rightarrow\frac{1}{\text{b}+1}=\frac{-\text{C}_\text{V}}{\text{R}}$
$\Rightarrow\text{b}+1=\frac{-\text{R}}{\text{C}_\text{V}}=\frac{-(\text{C}_\text{P}-\text{C}_\text{V})}{\text{C}_\text{V}}=-\gamma+1$
$\Rightarrow\text{b}=-\gamma$

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Question 43 Marks

In Joly's differential steam calorimeter, $3g$ of an ideal gas is contained in a rigid closed sphere at $20°C$. The sphere is heated by steam at $100°C$ and it is found that an extra $0.095g$ of steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in $Jg^{-1} K^{-1}$. The latent heat of vaporisation of water = $540cal-g^{-1}$.

Answer

In Joly’s differential steam calorimeter,$\text{C}_\text{v}=\frac{\text{m}_2\text{L}}{\text{m}_1(\theta_2-\theta_1)}$
$m_2$ = Mass of steam condensed = $0.095g, L = 540Cal/g = 540 × 4.2J/g m_1$ = Mass of gas present = 3g, $\theta_1=20^\circ\text{C},\ \theta_2=100^\circ\text{C}$
$\Rightarrow\text{C}_\text{v}=\frac{0.095\times540\times4,.2}{3(100-20)}=0.89\approx0.9\text{J/g-K}$

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Question 53 Marks

An amount Q of heat is added to a monatomic ideal gas in a process in which the gas performs a work $\frac{\text{Q}}{2}$ on its surrounding. Find the molar heat capacity for the process.

Answer

Q = Amt of heat given Work done $=\frac{\text{Q}}{2},$ $\triangle\text{Q}=\text{W}+\triangle\text{U}$ for monoatomic gas,$\Rightarrow\triangle\text{U}=\text{Q}-\frac{\text{Q}}{2}=\frac{\text{Q}}{2}$
$\Rightarrow\text{V}=\text{n}\frac{3}{2}\text{RT}=\frac{\text{Q}}{2}$
$=\text{nT}\times\frac{3}{2}\text{R}=3\text{R}\times\text{nT}$
Again Q = n CpdT Where $C_P$ > Molar heat capacity at const. pressure. $3RnT = ndTC_P \Rightarrow C_P = 3R$

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Question 63 Marks

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of R the gas for the process is given by $\text{C}=\text{C}_\text{v}+\frac{\text{R}}{2}.$

Answer

$\text{P}=\text{KV}$$\Rightarrow\frac{\text{nRT}}{\text{V}}=\text{KV}$
$\Rightarrow\text{RT}=\text{KV}^2$
$\Rightarrow\text{R}\triangle\text{T}=2\text{KV}\triangle\text{V}$
$\Rightarrow\triangle\text{V}=\frac{\text{R}\triangle\text{T}}{2\text{KV}}$
$\text{dQ}=\text{du}+\text{dw}$
$\Rightarrow\text{m}_\text{c}\text{dT}=\text{C}_\text{v}\text{dt}+\text{pdv}$
$\Rightarrow\text{m}_\text{c}\text{dT}=\text{C}_\text{v}\text{dt}+\frac{\text{pRdT}}{2\text{KV}}$
$\Rightarrow\text{m}_\text{c}=\text{C}_\text{v}+\frac{\text{RKV}}{\text{2KV}}$
$\Rightarrow\text{m}_\text{c}=\text{C}_\text{v}+\frac{\text{R}}{\text{2}}$

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Question 73 Marks

The specific heat capacities of hydrogen at constant volume and at constant pressure are $2.4 \mathrm{cal}^{-} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and $3.4 \mathrm{cal}-$ $\mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ respectively. The molecular weight of hydrogen is $2 \mathrm{~g}-\mathrm{mol}^{-1}$ and the gas constant, $\mathrm{R}=8.3 \times 10^7 \mathrm{erg}^{\circ} \mathrm{C}^{-1} \mathrm{~mol}^{-}$ ${ }^1$. Calculate the value of $J$.

Answer

$\mathrm{C}_{\mathrm{V}} \mathrm{H}_2=2.4 \mathrm{Cal} / \mathrm{g}^{\circ} \mathrm{C}, \mathrm{CpH}_2=3.4 \mathrm{Cal} / \mathrm{g}^{\circ} \mathrm{C}$
$\mathrm{M}=2 \mathrm{~g} / \mathrm{Mol}, \mathrm{R}=8.3 \times 10^7 \mathrm{erg} / \mathrm{mol}-{ }^{\circ} \mathrm{C}$
We know, $C_p-C_V=1 \mathrm{Cal} / \mathrm{g}^{\circ} \mathrm{C}$
So, difference of molar specific heats
$=C_p \times M-C_V \times M=1 \times 2=2 \mathrm{Cal} / g^{\circ} \mathrm{C}$
Now, $2 \times \mathrm{J}=\mathrm{R}$
$\Rightarrow 2 \times J=8.3 \times 10^7 \mathrm{erg} / \mathrm{mol}-{ }^{\circ} \mathrm{C}$
$\Rightarrow J=4.15 \times 10^7 \mathrm{erg} / \mathrm{cal} .$

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Question 83 Marks

A mixture contains 1 mole of helium ( $\left.C_p=2.5 R, C_v=1.5 R\right)$ and 1 mole of hydrogen $\left(C_p=3.5 R, C_v=2.5 R\right)$. Calculate the values of $\mathrm{C}_{\mathrm{p}}, \mathrm{C}_{\mathrm{v}}$ and $\gamma$ for the mixture.

Answer

$C p^{\prime}=2.5, R C p^{\prime \prime}=3.5 R C v^{\prime}=1.5 R, C v^{\prime \prime}=2.5 R n_1=n_2=1 \mathrm{~mol},\left(n_1+n_2\right) C_v d T=n_1 C_v d T+n_2 C_v d T$
$\Rightarrow\text{C}_\text{v}=\frac{\text{n}_1\text{Cv}'+\text{n}_2\text{Cv}''}{\text{n}_1+\text{n}_2}$
$=\frac{1.5\text{R}+2.5\text{R}}{2}2\text{R}$
$\Rightarrow\text{C}_\text{p}=\text{C}_\text{v}+\text{R}=2\text{R}+\text{R}=3\text{R}$
$\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{v}}=\frac{3\text{R}}{2\text{R}}=1.5$

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Question 93 Marks

Air $(\gamma=1.4)$ is pumped at 2atm pressure in a motor tyre at 20°C. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre? Neglect any mixing with the atmospheric air.

Answer

$\gamma=1.4, T_1 = 20^\circ C = 293k,P_1 = 2atm, P_2= 1atm$
We know for adiabatic process,
$\text{P}_1^{1-\gamma}\times\text{T}_1^\gamma=\text{P}_2^{1-\gamma}\times\text{T}_2^\gamma$
$\Rightarrow(2)^{1-1.4}\times(293)^{1.4}=(1)^{1-1.4}\times\text{T}_2^{1.4}$
$\Rightarrow(2)^{0.4}\times(293)^{1.4}=\text{T}_2^{1.4}$
$\Rightarrow2153.78=\text{T}_2^{1.4}$
$\Rightarrow\text{T}_2=(2153.78)^\frac{1}{1.4}=240.3\text{K}$

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Question 103 Marks

A sample of air weighing 1.18 g occupies $1.0 \times 10^3 \mathrm{~cm}^3$ when kept at 300 K and $1.0 \times 10^5 \mathrm{~Pa}$. When 2.0 cal of heat is added to it at constant volume, its temperature increases by $1^{\circ} \mathrm{C}$. Calculate the amount of heat needed to increase the temperature of air by $1^{\circ} \mathrm{C}$ at constant pressure if the mechanical equivalent of heat is $4.2 \times 10^7 \mathrm{erg} \mathrm{~cal}^{-1}$. Assume that air behaves as an ideal gas.

Answer

$m = 1.18g, V = 1 \times 10^3cm^3= 1L, T = 300k, P = 10^5Pa$
$\text{PV}=\text{nRT},\ \text{n}=\frac{\text{PV}}{\text{RT}}=10^5=\text{atm}$
$\text{N}=\frac{\text{PV}}{\text{RT}}=\frac{1}{8.2\times10^{-2}\times3\times10^2}$
$=\frac{1}{8.2\times3}=\frac{1}{24.6}$
Now, $\text{C}_\text{v}=\frac{1}{\text{n}}\times\frac{\text{Q}}{\text{dt}}=24.6\times2=49.2$$\text{C}_\text{p}=\text{R}+\text{C}_\text{v}=1.987+49.2=51.187$
$\text{Q}=\text{nC}_\text{p}\text{dT}=\frac{1}{24.6}\times51.187\times1=2.08\text{Cal}$

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Question 113 Marks

5 g of a gas is contained in a rigid container and is heated from $15^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. Specific heat capacity of the gas at constant volume is $0.172 \mathrm{cal} {-\mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}}$ and the mechanical equivalent of heat is $4.2 \mathrm{~J}-\mathrm{cal}^{-1}$. Calculate the change in the internal energy of the gas.

Answer

$\mathrm{m}=5 \mathrm{~g}, \Delta \mathrm{t}=25-15=10^{\circ} \mathrm{C}$
$\mathrm{C}_{\mathrm{V}}=0.172 \mathrm{cal} / \mathrm{g}-{ }^{\circ} \mathrm{CJ}=4.2 \mathrm{~J} / \mathrm{Cal}$
$\mathrm{dQ}=\mathrm{du}+\mathrm{dw}$
Now, $\mathrm{V}=0$ (for a rigid body)
So, $d w=0$
So, $d Q=d u$
$\mathrm{Q}=\mathrm{msdt}=5 \times 0.172 \times 10=8.6 \mathrm{cal}$
$=8.6 \times 4.2=36.12 \text { Joule }$

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