Online Test: Thermodynamics | Class 11 Physics

Question 1 of 18

The first law of thermodynamics is expressed as:

$\Delta Q=\Delta U-\Delta W$

$\Delta Q=\Delta U+\Delta W$

$\Delta U=\Delta Q+\Delta W$

$\Delta W=\Delta Q+\Delta U$

Explanation: Heat added equals the rise in internal energy plus the work done by the gas: $\Delta Q=\Delta U+\Delta W$.

Question 2 of 18

The zeroth law of thermodynamics defines the concept of:

entropy

temperature

work

enthalpy

Explanation: Thermal equilibrium between bodies via a third defines temperature.

Question 3 of 18

For an ideal gas, internal energy depends only on:

pressure

volume

temperature

shape of the container

Explanation: Internal energy of an ideal gas is a function of temperature alone.

Question 4 of 18

The work done by a gas is given by:

$W=\int V\,dP$

$W=\int P\,dV$

$W=P/V$

$W=PV$

Explanation: $W=\int P\,dV$, the area under the $P$–$V$ curve.

Question 5 of 18

In an isothermal process for an ideal gas, $\Delta U$ equals:

$\Delta Q$

$\Delta W$

zero

$Q_1-Q_2$

Explanation: Temperature is constant, so $\Delta U=0$.

Question 6 of 18

An adiabatic process obeys:

$PV=\text{const}$

$P/V=\text{const}$

$PV^\gamma=\text{const}$

$V/T=\text{const}$

Explanation: With $\Delta Q=0$, $PV^\gamma=\text{const}$ where $\gamma=\frac{C_p}{C_v}$.

Question 7 of 18

The work done in an isochoric process is:

maximum

$P\Delta V$

zero

$nRT\ln(V_2/V_1)$

Explanation: Volume is constant, so $dV=0$ and $W=0$.

Question 8 of 18

Mayer's relation is:

$C_p-C_v=R$

$C_p+C_v=R$

$C_p/C_v=R$

$C_v-C_p=R$

Explanation: $C_p-C_v=R$ for one mole of an ideal gas.

Question 9 of 18

The work done in an isothermal expansion from $V_1$ to $V_2$ is:

$P(V_2-V_1)$

$nRT\ln(V_2/V_1)$

$\frac{P_1V_1-P_2V_2}{\gamma-1}$

zero

Explanation: $W=\int\frac{nRT}{V}dV=nRT\ln(V_2/V_1)$.

Question 10 of 18

The efficiency of a Carnot engine between source $T_1$ and sink $T_2$ is:

$1-\frac{T_2}{T_1}$

$\frac{T_2}{T_1}$

$1-\frac{T_1}{T_2}$

$\frac{T_1}{T_2}$

Explanation: $\eta=1-\frac{T_2}{T_1}$ with temperatures in kelvin.

Question 11 of 18

A heat engine takes 1000 J and rejects 750 J per cycle. Its efficiency is:

75%

25%

50%

10%

Explanation: $\eta=1-\frac{750}{1000}=0.25=25\%$.

Question 12 of 18

The efficiency of any heat engine is always:

greater than 1

equal to 1

less than 1

exactly 0.5

Explanation: Some heat is always rejected, so $\eta=1-\frac{Q_2}{Q_1}<1$.

Question 13 of 18

For a Carnot engine the efficiency is 100% only if:

$T_1=T_2$

$T_2=0\ \text{K}$

$T_1=0\ \text{K}$

$Q_1=Q_2$

Explanation: $\eta=1-\frac{T_2}{T_1}=1$ needs $T_2=0$ K, which is unattainable.

Question 14 of 18

A Carnot engine works between 400 K and 300 K. Its efficiency is:

75%

25%

33%

57%

Explanation: $\eta=1-\frac{300}{400}=0.25=25\%$.

Question 15 of 18

The coefficient of performance of a refrigerator is:

$\frac{W}{Q_2}$

$\frac{Q_2}{W}$

$\frac{Q_1}{W}$

$1-\frac{Q_2}{Q_1}$

Explanation: $\beta=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}$.

Question 16 of 18

In a cyclic process the change in internal energy is:

maximum

zero

equal to the work done

negative

Explanation: After a full cycle the system returns to its initial state, so $\Delta U=0$.

Question 17 of 18

Heat supplied to a gas at constant volume goes entirely into:

work done

internal energy

expansion

heat loss

Explanation: $W=0$ at constant volume, so $\Delta Q=\Delta U=nC_v\Delta T$.

Question 18 of 18

On a $P$–$V$ diagram, the adiabatic curve compared with the isotherm through the same point is:

less steep

steeper

identical

horizontal

Explanation: Adiabatic slope is $\gamma$ times the isothermal slope, and $\gamma>1$, so it is steeper.

Online Test — Thermodynamics