3 Marks Question

Question 13 Marks

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Answer

(a) When the stopcock is opened suddenly, the volume of gas available at one air pressure doubles and the pressure will reduce to half i.e. 0.5 air pressure.
(b) There will be no change in the internal energy of the gas, because no work is done on or by the gas.
(c) If the gas is considered ideal then there will be no change in its temperature because it does not do any work in expansion. That means $\Delta T=0$.
(d) No, because the gas is expanding freely, the process cannot be controlled (rapid process), hence the gas will not attain equilibrium in the intermediate stages. Yes, after some time the gas will definitely achieve equilibrium and the state of the gas will be located on the P, V, T surface.

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Question 23 Marks

In changing the state of a gas adiabatically from an equilibrium state $A$ to another equilibrium state $B$, and amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? $($ Take $1 cal =4.19 J)$

Answer

Here the adiabatic process is
$\begin{aligned} \Delta Q & =0 \\ \Delta W & =-23.3 J\end{aligned}$
From first law of thermodynamics
$
\Delta Q=dU+\Delta W
$
$\begin{array}{lrl}\text { Put the value } & 0 & = dU -22.3 \\ \text { or } & \Delta U & =22.3 J\end{array}$
In another process
$
\begin{aligned}
\Delta Q & =9.35 \text { Calories } \\
& =9.35 \times 4.2 J=39.3 \text { Joule } \\
\Delta W & =?
\end{aligned}
$
$\therefore \quad \begin{aligned} \Delta Q & = dU +\Delta W \\ \Delta W & =\Delta Q -\Delta U \\ & =39.3-22.3 J \\ & =17.0 J\end{aligned}$

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Question 33 Marks

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer

No heat exchange is allowed. Hence the process is adiabatic. Now given for adiabatic transformation.
$\begin{array}{ll}\text { Let } & V _1=x \\ \therefore & V_2=\frac{1}{2} \quad V_1=\frac{1}{2} x\end{array}$
$\gamma=\frac{ C _{ p }}{ C _{ V }}$
For diatomic gass or hydrogen
$\begin{aligned} \gamma & =\frac{7}{5}=1.4 \\ \frac{ P _2}{ P _1} & =? \\ P _1 V_1^\gamma & = P _2 V_2^\gamma\end{aligned}$
or
$
\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma=\left(\frac{x}{x / 2}\right)^{1.4}=(2)^{1.4}
$
$\begin{array}{ll}\therefore & \frac{ P _2}{ P _1}=2.64 \\ & P _2=2.64 P _1\end{array}$
Therefore, the pressure will have to be 2.64 times more than the initial pressure.

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Question 43 Marks

What amount of heat must be supplied to $2.0 \times 10^{-2} kg$ of nitrogen (at room temperature) to raise its temperature by $45^{\circ} C$ at constant pressure? (Molecular mass of $N _2=28 ; R =8.3 J mol ^{-1} K^{-1}$ ).

Answer

Given that
mass of gas
$
\begin{aligned}
m & =2 \times 10^{-2} kg \\
& =2 \times 10^{-2} \times 10^3 g=20 g
\end{aligned}
$
increasing in temperature
$
\Delta T=45^{\circ} C
$
molecular weight of $N _2=28 g$
$n$ moles in $m$ gram gas then
$
n=\frac{m}{M}=\frac{20}{28}=\frac{5}{7}
$
$
R=8.3 J mol^{-1} K^{-1}
$
Since nitrogen is a diatomic gas, the value of specific heat of the gas at constant pressure is
$
C_{P}=\frac{7}{2} R=\frac{7}{2} \times 8.3 J mol^{-1} K^{-1}
$
Let need of energy
$
\begin{aligned}
\Delta Q & =? \\
\therefore \quad \Delta Q & =nC_{p} \Delta T \\
& =\frac{5}{7} \times \frac{7}{2} \times 8.3 \times 45 J \\
& =933.75 J \\
& \simeq 934 J
\end{aligned}
$

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Question 53 Marks

P-V diagram of a gas is shown in the attached figure. How much work will have to be done to move the gas from point $A$ to point $B~?$

Answer

The process from A to B is at constant pressure, hence the work done.
$\begin{aligned}W & =\int_{V_1}^{V_2} P d V \\& =P \int_{V_{A}}^{V_{B}} d V=P[V]_{V_{A}}^{V_{B}}=P\left(V_{B}-V_{A}\right)\end{aligned}$
Here $P =5$ Netwon $/ m ^2, V_{ A }=2 m^3$
$V_{B}=6 m^3$
Work $W =5 \times(6-2)=5 \times 4=20$ Joule

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Question 63 Marks

Write the main points of efficiency of Carnot engine.

Answer

Main points of efficiency :
(i) Efficiency does not depend on the working substance.
(ii) Efficiency is greater when the temperature difference between source and sink is greater. It occurs, but it is always less than $100 \%$.
(iii) The efficiency of the Carnot engine depend only on the temperature of the source and the sink
(iv) All reversible (ideal) engines working between two equal temperature have the same efficiency.

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Question 73 Marks

As gas expands from initial volume $V_1$ to $V_2$ in three process whose $P$$-V$ curve is shown in the figure, which process is which? In which of the following processes is the gas used the most?

Answer

In the given figure, isobaric process AB , isothermal process AC and adiabatic process AD . In these the are enclosed between the isobaric process $A B$ and the volume axis is maximum. Therefore most work will be done in this process.

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Question 83 Marks

The output power of the motor of a refrigerator is 240 W . The temperature of the freezing chamber is 240 K and the outside air is 300 K . How much heat can be removed from the freezer in 10 minutes? Assuming that there is ideal efficiency, what is the minimum time in which 10 kg of water at 273 K can be converted in to ice? $J =4.2 \times 10^3 JK cal ^{-1}$.

Answer

We know that :
$\alpha=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}$
Here given :
$\begin{aligned}T_1 & =300 K \\T_2 & =240 K \\W & =240 W=240 J / s \\Q_1 & =? \\t & =?\end{aligned}$
$\begin{aligned}\therefore \quad Q_2 & =W(\alpha)=\frac{W \cdot T_2}{T_1-T_2} \\& =240 \times \frac{240}{300-240} \\& =240 \times \frac{240}{60}=240 \times 4=960 J / 5\end{aligned}$
(i) Suppose heat removed from freezing chamber in 10 minutes $= Q$.
$\begin{aligned}\therefore \quad Q & =Q_2 \times 10 min . \\& =960 \times 10 \times 60=576000 J \\Q & =576000 J \\& =\frac{576000}{4.2 \times 10^3}=137.14 Kcal \\& {\left[\because J=4.2 \times 10^3 J / K Cal\right] }\end{aligned}$
(ii) To convert 1 kg of water into ice at 273 K desired heat
$\begin{aligned}Q & =m \times L=1 \times 80 Kcal \\& =80 \times 4.2 \times 10^3 J\end{aligned}$
Suppose the heat extracted in time t is $Q ^{\prime}$.
$\therefore$ Rate of heat removal from freezing chamber
$=\frac{80 \times 4.2 \times 10^3}{t} J / s$
This rate should be equal to $Q_2$
i.e.
$\begin{aligned}960 & =\frac{80 \times 4.2 \times 10^3}{t} \\t & =\frac{80 \times 4.2 \times 10^3}{960} \\& =\frac{8 \times 42}{96} \times 10^2=\frac{336}{96} \times 10^2 \\t & =3.5 \times 10^2{ }_5=350s\end{aligned}$

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Question 93 Marks

By suddenly compressing a gas at a temperature of 300 K, its pressure is made 8 times the initial pressure. Calculate the temperature increase due to compression $(\gamma=1.5)$.

Answer

$\begin{array}{l}T_1=300 K \\\text {Assume}\quad P_1=P \text { and } P_2=8 P\end{array}$
The increase in temperature due to compression is assumed to be $T_2$ is the adiabatic process.
$\begin{array}{ll}\therefore & P_1^{1-\gamma} T_1^\gamma=P_2^{1-\gamma} T_2^\gamma \text { constant } \\& \left(\frac{P_1}{P_2}\right)^{1-\gamma}=\left(\frac{T_2}{T_1}\right)^\gamma\end{array}$
Putting the value :
$\begin{array}{l}\left(\frac{P_1}{P_2}\right)^{\frac{1-\gamma}{\gamma}}=\frac{T_2}{T_1} \\\left(\frac{P}{8 P}\right)^{\frac{1-1.5}{1.5}}=\left(\frac{T_2}{300}\right) \\\left(\frac{1}{8}\right)^{\frac{-1}{3}}=\frac{T_2}{300} \\\left(\frac{1}{2}\right)^{-1}=\frac{T_2}{300} \Rightarrow 2=\frac{T_2}{300} \\\therefore T_2=600 K
\end{array}$
Increase in temperature due to compression
$\begin{array}{l}=T_2-T_1 \\=600 K-300 K \\=300 K\end{array}$

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Question 103 Marks

The pressure of the wheel of motor car at any instant is equal to 2 atmospheric pressure at the temperature is $15^{\circ} C$. At this moment the tube of the wheel bursts. Find the decrease in temperature of the released $\operatorname{air}(\gamma=1.4)$

Answer

Given that :
$P_1=2 Atmospheric$
$\begin{aligned}T_1 & =15^{\circ} C=15+273=288 K \\\gamma & =1.4 \\T_2 & =? \\P_2 & =1 \text { Atmospheric }\end{aligned}$
And
We know that
\begin{array}{l}

P_1^{1-\gamma} T_1^\gamma=P_2^{1-\gamma} T_2^\gamma=\text { Constant } \\\begin{aligned}\left(\frac{P_1}{P_2}\right)^{1-\gamma} & =\left(\frac{T_2}{T_1}\right)^\gamma \\\left(\frac{T_2}{T_1}\right)^\gamma & =\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \\T_2 & =T_1\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \\& =288\left(\frac{1}{2}\right)^{\frac{1.4-1}{1.4}} \\& =288\left(\frac{1}{2}\right)^{2 / 7} \\

T_2 & =\frac{280}{2^{2 / 7}}\end{aligned}\end{array}
Taking log both sides :
$\begin{aligned}\log T_2 & =\log 288-\frac{2}{7}\log 2 \\& =2.4594-\frac{2}{7}(0.3010) \\& =2.4594-0.086\\\log T_2 & =2.3734 \\ T_2 & =\text { Anti } \log 2.3734\\\therefore\quad T_2 & =236.3 K\end{aligned}$
Decrease in temperature
$\begin{array}{l}=288-236.3 \\=51.7 K\end{array}$

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Question 113 Marks

The efficiency of an ideal engine is $\frac{1}{8}$ access. After reducing the temperature 100 K , increase to $\frac{1}{6}$ of access. Find the original and final temperature.

Answer

$ \eta_1=\frac{1}{8}$ From position first
$\eta_2=\frac{1}{6}$ From position second
Suppose $\quad T_1=$ Temperature of source
$T _2=$ Original temperature of access
$T _2^{\prime}=$ Final temperature of access
$=T_2-100$
$\therefore$ Relation $\eta_1=1-\frac{T_2}{T_1}$
From position 1st
$\frac{1}{8}=1-\frac{T_2}{T_1}$
or $\quad \frac{ T _2}{T_1}=1-\frac{1}{8}=\frac{7}{8}\ldots\ldots (1)$
From position 2nd
$\begin{array}{l}\frac{1}{6}=1-\frac{ T _2^{\prime}}{ T _1}=1-\left(\frac{\left( T _2-100\right)}{ T _1}\right) \\ \frac{1}{6}=1-\frac{ T _2}{T_1}+\frac{100}{T_1}\ldots\ldots (2)\end{array}$
Putting values from (1) to (2) gives
$\begin{aligned} \frac{1}{6} & =1-\frac{7}{8}+\frac{100}{T_1} \\\text {or }\quad \frac{1}{6} & =\frac{1}{8}+\frac{100}{T_1} \\ \frac{1}{6}-\frac{1}{8} & =\frac{100}{T_1} \text { or } \frac{1}{24}=\frac{100}{T_1} \\\text {or}\quad T_1 & =2400 K\end{aligned}$
Putting the value in equation (1)
$\begin{aligned} \frac{ T _2}{2400} & =\frac{7}{8} \\\text {or}\quad T_2 & =\frac{7 \times 2400}{8}=7 \times 300=2100 K \\\text {Therefore}\quad T _2^{\prime} & = T _2-100 \\ & =2100-100=2000 K\end{aligned}$

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Physics 3 Marks Question

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