MCQ 11 Mark
In the usual notation, the isothermal work, $W =$
- A
$P\left(V_f-V_i\right)$
- B
$n R T\left(P_i / P_f\right)$
- ✓
$n R T \ln \left(P_i / P_f\right)$
- D
$n R T\left(P_f / P_i\right)$.
AnswerCorrect option: C. $n R T \ln \left(P_i / P_f\right)$
$n R T \ln \left(P_i / P_f\right)$
View full question & answer→MCQ 21 Mark
In an isobaric process, in the usual notation,
- A
$W={ }_n C_V\left(T_f-T_i\right)$
- ✓
$Q={ }_n C_p\left(T_f-T_i\right)$
- C
$\Delta U=n R\left(T_f-T_i\right)$
- D
$W =0$.
AnswerCorrect option: B. $Q={ }_n C_p\left(T_f-T_i\right)$
$Q={ }_n C_p\left(T_f-T_i\right)$
View full question & answer→MCQ 31 Mark
In an isothermal process, in the usual notation,
- A
$W=P\left(V_f-V_i\right)$
- B
$W=0$
- C
$W=V\left(P_f-P_i\right)$
- ✓
$W=n R T \ln \left(V_f / N_i\right)$.
AnswerCorrect option: D. $W=n R T \ln \left(V_f / N_i\right)$.
$W=n R T \ln \left(V_f / N_i\right)$.
View full question & answer→MCQ 41 Mark
In an adiabatic process, in the usual notation,
- A
$\Delta P=0$
- B
$\Delta V=0$
- ✓
$Q=0$
- D
$\Delta U=0$.
View full question & answer→MCQ 51 Mark
In an isobaric process, in the usual notation,
AnswerCorrect option: A. $W=P\left(V_f-V_i\right)$
$W=P\left(V_f-V_i\right)$
View full question & answer→MCQ 61 Mark
In an adiabatic process, in the usual notation,
- A
$TV ^{ γ}=$ constant
- B
$PT \gamma=$ constant
- C
$W =0$
- ✓
$PV ^γ=$ constant.
AnswerCorrect option: D. $PV ^γ=$ constant.
$PV ^\gamma=$ constant.
View full question & answer→MCQ 71 Mark
In an isothermal process, in the usual notation,
MCQ 81 Mark
The adiabatic constant $\gamma$ for argan is
- A
$4 / 3$
- B
$7 / 5$
- C
$6 / 5$
- ✓
$5 / 3$.
AnswerCorrect option: D. $5 / 3$.
$5 / 3$.
View full question & answer→MCQ 91 Mark
The internal energy of one mole of nitrogen at $300 K$ is about $6225 J$. Its internal energy at $400 K$ will be about
AnswerCorrect option: A. $8300 J$
$8300 J$
View full question & answer→MCQ 101 Mark
The internal energy of one mole of oxygen is
- A
$\frac{5}{2} R T$
- B
$5 RT$
- ✓
$\frac{3}{2} R T$
- D
AnswerCorrect option: C. $\frac{3}{2} R T$
$\frac{3}{2} R T$
View full question & answer→MCQ 111 Mark
The internal energy of one mole of organ is
- A
$\frac{5}{2} R T$
- B
- ✓
$\frac{3}{2} R T$
- D
AnswerCorrect option: C. $\frac{3}{2} R T$
$\frac{3}{2} R T$
View full question & answer→MCQ 121 Mark
The coefficient of performance of a Carnot refrigerator working between $T_H$ and $T_C$ is $K$ and the efficiency of a Carnot engine working between the same $T_H$ and $T_c$ is $\eta$. Then
AnswerCorrect option: D. $\eta k =\frac{Q_c}{Q_H}$
$\eta k =\frac{Q_c}{Q_H}$
View full question & answer→MCQ 131 Mark
The coefficient of performance of a Carnot refrigerator is 4. If $T _{ C }=250 K$, then $T _H=$
- A
$625 K$
- B
$310 K$
- ✓
$312.5 K$
- D
$320 K$
AnswerCorrect option: C. $312.5 K$
$312.5 K$
View full question & answer→MCQ 141 Mark
In an isothermal process, in the usual notation,
- ✓
$PV =$ constant
- B
$V / T =$ constant
- C
$P / T=$ constant
- D
$Q =0$.
AnswerCorrect option: A. $PV =$ constant
$PV =$ constant
View full question & answer→MCQ 151 Mark
If a Carnot refrigerator works between $250 K$ and $300 K$, its coefficient of performance $=$
View full question & answer→MCQ 161 Mark
If a Carnot engine receives $5000 J$ from a hot reservoir and rejects $4000 J$ to a cold reservoir, the efficiency of the engine is
- A
$25 \%$
- B
$10 \%$
- C
$1 / 9$
- ✓
$20 \%$.
AnswerCorrect option: D. $20 \%$.
$20 \%$.
View full question & answer→MCQ 171 Mark
The efficiency of a Carnot engine working between $T_H=400 K$ and $T_C=300 K$ is
- A
$75 \%$
- ✓
$25 \%$
- C
$1 / 3$
- D
$4 / 7$
AnswerCorrect option: B. $25 \%$
$25 \%$
View full question & answer→MCQ 181 Mark
The coefficient of performance of a Carnot refrigerator is given by $K =$
- ✓
$T_c\left(T_H-T_c\right)$
- B
$\left(T_H-T_c\right) / T_C$
- C
$T_H /\left(T_H-T_C\right)$
- D
$\left(T_H-T_c\right) / T_H$.
AnswerCorrect option: A. $T_c\left(T_H-T_c\right)$
$T_c\left(T_H-T_c\right)$
View full question & answer→MCQ 191 Mark
The efficiency of a Carnot engine is given by $K=$
- ✓
$T_c /\left(T_H-T_C\right)$
- B
$\left(T_H-T_c\right) / T_C$
- C
$T_H /\left(T_H-T_C\right)$
- D
$\left(T_H-T_C\right) / T_H$
AnswerCorrect option: A. $T_c /\left(T_H-T_C\right)$
$T_c /\left(T_H-T_c\right)$
View full question & answer→MCQ 201 Mark
In a cyclic process, the area enclosed by the loop in the $P - V$ plane corresponds to
- A
$\Delta U$
- ✓
$W$
- C
$Q-W$
- D
$W - Q$.
View full question & answer→MCQ 211 Mark
The efficiency of a heat engine is given by $\eta=$
- A
$Q_H / W$
- B
$W / Q_c$
- ✓
$W / Q _{ H }$
- D
$Q_c / W$.
AnswerCorrect option: C. $W / Q _{ H }$
$W / Q _{ H }$
View full question & answer→MCQ 221 Mark
- A
$\Delta U=Q$
- B
$Q=0$
- C
$W=0$
- ✓
$W=Q$
View full question & answer→MCQ 231 Mark
In an adiabatic process, in the usual notation, $W=$
- ✓
$\frac{P_{ i } V_{ i }-P_{ f } V_{ f }}{\gamma-1}$
- B
$\frac{n R \Delta T}{\gamma}$
- C
$n R \gamma\left(T_{ i }-T_{ f }\right)$
- D
$\frac{P_i V_i-P_f V_f}{\gamma}$.
AnswerCorrect option: A. $\frac{P_{ i } V_{ i }-P_{ f } V_{ f }}{\gamma-1}$
$\frac{P_1 V_1-P_{ f } V_{ f }}{\gamma-1}$
View full question & answer→MCQ 241 Mark
If $Q$ and $\Delta u$ are expressed in cal and $W$ is expressed in joule, then,
AnswerCorrect option: D. $Q =\Delta U +( W / J )$
$Q=\Delta U+(W / J)$
View full question & answer→MCQ 251 Mark
According to the first law of thermodynamics, in the usual notation,
- ✓
$Q =\Delta U + W$
- B
$Q=\Delta U-W$
- C
$Q=W-\Delta U$
- D
$Q=-(\Delta U+W)$.
AnswerCorrect option: A. $Q =\Delta U + W$
$Q=\Delta U+W$
View full question & answer→MCQ 261 Mark
During refrigeration cycle, heat is rejected by the refrigerant in the :
Answercondenser [See the textbook]
View full question & answer→MCQ 271 Mark
The second law of thermodynamics deals with transfer of:
MCQ 281 Mark
Efficiency of a Carnot engine is large when
AnswerCorrect option: A. $T _{ H }$ is large
$T _{ H }$ is large
$T_C$ is low
$T _{ H }- T _{ C }$ is large
$\left[\eta=\frac{T_{ H }-T_{ c }}{T_{ H }}=1-\frac{T_{ c }}{T_{ H }}\right]$
View full question & answer→MCQ 291 Mark
Which of the following is an example of the first law of thermodynamics?
- A
The specific heat of an object explains how easily it changes temperatures.
- ✓
While melting, an ice cube remains at the same temperature.
- C
When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
- D
After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
AnswerCorrect option: B. While melting, an ice cube remains at the same temperature.
While melting, an ice cube remains at the same temperature. [Here, ∆u = 0, W = Q]
When a refrigerator is unplugged, everything inside of it returns to room temperature after some time.
After falling down the hill, a ball’s kinetic energy plus heat energy equals the initial potential energy.
View full question & answer→MCQ 301 Mark
A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in the internal energy of the system?
MCQ 311 Mark
A gas at N.T.P. is suddenly compressed to $\left(\frac{1}{4}\right)^{ th }$ of its original volume. The final pressure in (Given $\gamma=$ ratio of specific heats $=\frac{3}{2}$ ) atmosphere is ( $P=$ original pressure)
- A
$4 P$
- B
$\frac{3}{2} P$
- ✓
$8 P$
- D
$\frac{1}{4} P$
Answer(c): Given, $V_1=V, V_2=\frac{V}{4}, \gamma=\frac{3}{2}$
$
\begin{aligned}
& P_1=P, P_2=? \\
& P_1 V_1^\gamma=P_2 V_2^\gamma \quad \text { (As process is adiabatic) } \\
& P\left(\frac{V \times 4}{V}\right)^{3 / 2}
=P_2 ; P_2=P \times 2^{2 \times 3 / 2}=8 P
\end{aligned}$
View full question & answer→MCQ 321 Mark
The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes adiabatically from $(P, \rho)$ to $\left(P^{\prime}, r^{\prime}\right)$. If $\frac{\rho^{\prime}}{\rho}=32$ then $\frac{P^{\prime}}{P}$ should be
Answer(b) : Given, $\frac{\rho}{\rho}=32, \gamma=\frac{\gamma}{5} ; \frac{\rho}{\rho^{\prime}}=\frac{V}{V}$. . . . .(i)
$P^{\prime} V^\gamma=P V^\gamma$
(for adiabatic process)
$
\frac{P^{\prime}}{P}=\left(\frac{V}{V^{\prime}}\right)^\gamma=\left(\frac{\rho^{\prime}}{\rho}\right)^\gamma=(32)^{7 / 5}=2^7=128
$
View full question & answer→MCQ 331 Mark
A gas is compressed at a constant pressure of $50\ N / m ^2$ from a volume of $10 m ^3$ to a volume of $4 m ^3$. Energy of $100 J$ is then added to the gas by heating. Its internal energy is
- ✓
increased by $400 J$
- B
increased by $200 J$
- C
increased by $100 J$
- D
decreased by $200 J$
AnswerCorrect option: A. increased by $400 J$
$P=50 N / m ^2, V_1=10 m ^3, V_2=4 m ^3, Q=100 J$
$\Delta W=-P d V=-50(10-4)=-300 J$
$\Delta Q=\Delta U+\Delta W+100=\Delta U-300$
$\Rightarrow \Delta U=+400 J $
View full question & answer→MCQ 341 Mark
A gas at N.T.P. is suddenly compressed to one-fourth of its original volume. If $\gamma=1.5$, then the final pressure is
Answer(c): As, $V_1=V, V_2=V / 4, \gamma=1.5, P_1=P, P_2=$ ?
Process is adiabatic
$
\begin{aligned}
\therefore & P_1 V_1^\gamma=P_2 V_2^\gamma \\
& P_2=P\left(\frac{V_1}{V_2}\right)^\gamma=P\left(\frac{V \times 4}{V}\right)^{1.5} \Rightarrow P_2=P(2)^{2 \times 1.5}=8 P
\end{aligned}
$
View full question & answer→MCQ 351 Mark
In a thermodynamic process, there is no exchange of heat between the system and surroundings. Then the thermodynamic process is
Answer(d) : In adiabatic process, no heat exchange between system and surrounding takes place.
View full question & answer→MCQ 361 Mark
An ideal refrigerator has freezer at a temperature of $-13^{\circ} C$. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is
- A
$320^{\circ} C$
- ✓
$39^{\circ} C$
- C
$325 K$
- D
$325^{\circ} C$
AnswerCorrect option: B. $39^{\circ} C$
(b) : Given, $\beta=5, T_2=-13^{\circ} C =-13+273=260 K$
$
\begin{aligned}
& \beta=\frac{T_2}{T_1-T_2} ; 5=\frac{260}{T_1-260} \\
& T_1-260=52 \\
& T_1=(260+52) K \\
& T_1=(260+52-273)^{\circ} C =39^{\circ} C
\end{aligned}
$
View full question & answer→MCQ 371 Mark
The efficiency of a heat engine is ' $\eta$ ' and the coefficient of performance of a refrigerator is ' $\beta$ '. Then
- A
$\eta=\frac{1}{\beta}$
- ✓
$\eta=\frac{1}{\beta+1}$
- C
$\eta \beta=\frac{1}{2}$
- D
$\eta=\frac{1}{\beta-1}$
AnswerCorrect option: B. $\eta=\frac{1}{\beta+1}$
(b) : $\eta=1-\frac{T_2}{T_1}, \beta=\frac{I_2}{T_1-T_2}$
So, $\frac{T_2}{T_1}=1-\eta, \beta=\frac{\frac{T_2}{T_1}}{1-\frac{T_2}{T_1}} ; \beta=\frac{1-\eta}{\eta}$
$
\begin{aligned}
& \beta \eta=1-\eta \\
& \eta(1+\beta)=1 \Rightarrow \eta=\frac{1}{1+\beta}
\end{aligned}
$
View full question & answer→MCQ 381 Mark
In a thermodynamic system $\Delta U$ represents the increase in its internal energy and $d W$ is the work done by the system then correct statement out of the following is
- A
$\Delta U=d W$ is an isothermal process
- ✓
$\Delta U=-d W$ is an adiabatic process
- C
$\Delta U=-d W$ is an isothermal process
- D
$\Delta U=d W$ is an adiabatic process.
AnswerCorrect option: B. $\Delta U=-d W$ is an adiabatic process
(b) : First law of thermodynamics, $d Q =d U+d W$
For isothermal process, $d U=0$
$
\Rightarrow d Q=d W
$
For adiabatic process, $d Q=0$
$
\Rightarrow \quad d U=-d W
$
View full question & answer→MCQ 391 Mark
A thermodynamic process of uncontrolled change satisfying the equation $Q=W=0$, is $[ Q =$ heat supplied, $W=$ work done $]$
Answer(d) : As $Q=0 ; W=0$ so there is no-work done on the system or by the system. And hence this thermodynamic process is free expansion.
View full question & answer→MCQ 401 Mark
A monoatomic ideal gas, initially at temperature $T_1$ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing the piston suddenly. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively then $T_1 / T_2$ is given by
- ✓
$\left(\frac{L_2}{L_1}\right)^{2 / 3}$
- B
$\left(\frac{L_1}{L_2}\right)$
- C
$\left(\frac{L_1}{L_2}\right)^{2 / 3}$
- D
$\left(\frac{L_2}{L_1}\right)$
AnswerCorrect option: A. $\left(\frac{L_2}{L_1}\right)^{2 / 3}$
(a) : For an ideal monatomic gas, the number of degrees of freedom, $f=3$
$
\gamma=1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}
$
In an adiabatic process, $T_1 V_1^\gamma V^{-1}=T_2 V_2^\zeta V^{-1}$
$
\begin{aligned}
& \Rightarrow \frac{T_1}{T_2}=\left(\frac{V_2}{V_1}\right)^{\gamma-1}=\left(\frac{V_2}{V_1}\right)^{\frac{5}{3}-1}=\left(\frac{V_2}{V_1}\right)^{\frac{2}{3}} \\
& \Rightarrow \frac{T_1}{T_2}=\left(\frac{V_2}{V_1}\right)^{\frac{2}{3}}=\left(\frac{L_2}{L_1}\right)^{\frac{2}{3}}
\end{aligned}
$
As volume is proportional to length.
View full question & answer→MCQ 411 Mark
In a adiabatic process, relation between $\Delta U$ and $\Delta W$ is
AnswerCorrect option: B. $\Delta U=-\Delta W$
(b) : In an adiabatic process, $\Delta Q=0$.
So, $\Delta U=-\Delta W$
View full question & answer→MCQ 421 Mark
If $\alpha$ is the coefficient of performance of a refrigerator and ' $Q_1$ ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' $Q_2$ ' is
- A
$\frac{\alpha Q_1}{\alpha-1}$
- B
$\frac{\alpha-1}{\alpha} Q_1$
- ✓
$\frac{\alpha \cdot Q_1}{1+\alpha}$
- D
$\frac{1+\alpha}{\alpha} Q_1$
AnswerCorrect option: C. $\frac{\alpha \cdot Q_1}{1+\alpha}$
(c) : Coefficient of performance of a refrigerator,
$
\begin{aligned}
& \alpha=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2} \\
& \Rightarrow \frac{1}{\alpha}=\frac{Q_1-Q_2}{Q_2}=\frac{Q_1}{Q_2}-1 \\
& \frac{1}{\alpha}+1=\frac{Q_1}{Q_2} \Rightarrow \frac{1+\alpha}{\alpha}=\frac{Q_1}{Q_2} \\
& \text { or } Q_2=\frac{Q_1 \alpha}{(1+\alpha)}
\end{aligned}
$
View full question & answer→