IMO Practice Test: Thermodynamics | Class 11 Physics

Question 1 of 12

A gas does 50 J of work in an isothermal expansion and 40 J in an adiabatic expansion between the same volumes. In the adiabatic case the internal energy change is:

$+40\ \text{J}$

$-40\ \text{J}$

$0$

$-50\ \text{J}$

Explanation: Adiabatic: $\Delta Q=0$, so $\Delta U=-\Delta W=-40\ \text{J}$.

Question 2 of 12

A monatomic gas ($\gamma=5/3$) is compressed adiabatically to one-eighth of its volume. The temperature becomes (from $T$):

$2T$

$4T$

$8T$

$\frac{T}{4}$

Explanation: $T_2=T(V_1/V_2)^{\gamma-1}=T\cdot8^{2/3}=4T$.

Question 3 of 12

The slope of the adiabatic curve to the isothermal curve at a point is in the ratio:

$1:\gamma$

$\gamma:1$

$1:1$

$\gamma^2:1$

Explanation: Adiabatic slope $=\gamma P/V$, isothermal slope $=P/V$, ratio $\gamma:1$.

Question 4 of 12

A Carnot engine has efficiency 40%. If the sink temperature is lowered by 50 K keeping the source fixed, the efficiency becomes 50%. The source temperature is:

$500\ \text{K}$

$400\ \text{K}$

$600\ \text{K}$

$250\ \text{K}$

Explanation: $T_2/T_1=0.6$ and $(T_2-50)/T_1=0.5$, so $50/T_1=0.1$, giving $T_1=500\ \text{K}$.

Question 5 of 12

A Carnot engine and a Carnot refrigerator work between the same two temperatures. If the engine's efficiency is $\eta$, the refrigerator's COP is:

$\eta$

$\frac{1}{\eta}$

$\frac{1-\eta}{\eta}$

$1-\eta$

Explanation: $\eta=1-T_2/T_1$ and $\beta=\frac{T_2}{T_1-T_2}=\frac{1-\eta}{\eta}$.

Question 6 of 12

For one mole of an ideal gas, $C_v=\frac{5}{2}R$. The value of $\gamma$ is:

$1.67$

$1.40$

$1.33$

$1.50$

Explanation: $C_p=C_v+R=\frac{7}{2}R$, so $\gamma=\frac{7/2}{5/2}=1.4$.

Question 7 of 12

Heat $Q$ is supplied to a diatomic gas ($\gamma=1.4$) at constant pressure. The fraction used to do work is:

$\frac{2}{7}$

$\frac{5}{7}$

$\frac{1}{2}$

$\frac{2}{5}$

Explanation: $\frac{W}{Q}=\frac{R}{C_p}=\frac{\gamma-1}{\gamma}=\frac{0.4}{1.4}=\frac{2}{7}$.

Question 8 of 12

An ideal refrigerator transfers 600 J of heat to a 300 K room using 100 J of work. The temperature of the cold interior is:

$250\ \text{K}$

$240\ \text{K}$

$260\ \text{K}$

$200\ \text{K}$

Explanation: $Q_2=Q_1-W=500\ \text{J}$; $\beta=Q_2/W=5=\frac{T_2}{T_1-T_2}$, so $T_2=5(300-T_2)\Rightarrow T_2=250\ \text{K}$.

Question 9 of 12

In a cyclic process the area enclosed by the loop on a $P$–$V$ diagram represents:

the change in internal energy

the net heat absorbed and net work done

the entropy change

zero

Explanation: For a cycle $\Delta U=0$, so net heat = net work = area of the loop.

Question 10 of 12

Two Carnot engines operate in series: the first between 800 K and $T$, the second between $T$ and 200 K, with equal efficiencies. Then $T$ is:

$500\ \text{K}$

$400\ \text{K}$

$600\ \text{K}$

$300\ \text{K}$

Explanation: Equal efficiency means $\frac{T}{800}=\frac{200}{T}$, so $T^2=160000$, $T=400\ \text{K}$.

Question 11 of 12

It is impossible to convert heat completely into work in a cyclic process. This is a statement of:

the zeroth law

the first law

the second law

Mayer's relation

Explanation: This is the Kelvin–Planck form of the second law of thermodynamics.

Question 12 of 12

A gas expands so that $PV^{1.4}=\text{const}$ and also is thermally insulated. If its volume doubles, the pressure becomes (from $P$):

$P/2$

$P\cdot2^{1.4}$

$P/2^{1.4}$

$2P$

Explanation: $P_2=P_1(V_1/V_2)^{1.4}=P/2^{1.4}$.

IMO Practice Test — Thermodynamics