Grade 11 : Physics - Thermodynamics (Set 4)

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Grade 11 : Physics - Thermodynamics (Set 4)— Questions & Detailed Solutions

System $A$ is in thermal equilibrium with System $B$, and System $B$ is separately in thermal equilibrium with System $C$.
According to the Zeroth Law of Thermodynamics, what can be concluded about Systems $A$ and $C$?

(A)

System A has greater thermal energy than System C

(B)

Heat flows spontaneously from System A to System C

(C)

System A and System C are in thermal equilibrium with each other

(D)

System A and System C have zero internal energy

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation: The Zeroth Law of Thermodynamics states that if two systems ($A$ and $C$) are each in thermal equilibrium with a third system ($B$), then they are in thermal equilibrium with each other. This implies they are at the same temperature, so $T_A = T_B = T_C$.

A gas expands isobarically at a constant pressure of $2.5 \times 10^5\text{ Pa}$ from an initial volume of $0.02m^3$ to a final volume of $0.06m^3$. What is the work done by the gas during this process?

(A)

5.0 kJ

(B)

10.0 kJ

(C)

15.0 kJ

(D)

20.0 kJ

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation: Work done in an isobaric process is given by $W = P \Delta V = P(V_2 - V_1)$.
$$W = (2.5 \times 10^5\text{ Pa})(0.06\text{ m}^3 - 0.02\text{ m}^3) = 2.5 \times 10^5 \times 0.04 = 10,000\text{ J} = 10.0\text{ kJ}$$

During a thermodynamic process, 500 J of heat is absorbed by a system while the system performs 180 J of work on its surroundings. What is the change in internal energy ($ΔU$) of the system?

(A)

+320 J

(B)

-320 J

(C)

+680 J

(D)

-680 J

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✓ Correct Option: (A)

Explanation: According to the First Law of Thermodynamics, $\Delta U = Q - W$.
Given $Q = +500\text{ J}$ (absorbed) and $W = +180\text{ J}$ (done by system):
$$\Delta U = 500\text{ J} - 180\text{ J} = +320\text{ J}$$

Process Type Condition / Constraint
Isochoric Volume remains constant ($ΔV = 0$)
Isobaric Pressure remains constant ($ΔP = 0$)
Isothermal Temperature remains constant ($ΔT = 0$)
Adiabatic No heat transfer ($Q = 0$)
Based on the table above, in which process is all heat supplied to an ideal gas entirely converted into work done by the gas?

Process Type	Condition / Constraint
Isochoric	Volume remains constant ($ΔV = 0$)
Isobaric	Pressure remains constant ($ΔP = 0$)
Isothermal	Temperature remains constant ($ΔT = 0$)
Adiabatic	No heat transfer ($Q = 0$)

(A)

Isochoric process

(B)

Isobaric process

(C)

Adiabatic process

(D)

Isothermal process

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation: For an ideal gas in an isothermal process, temperature is constant, so $\Delta U = 0$. By the First Law ($ΔU = Q - W$), we get $0 = Q - W \implies Q = W$. Thus, all added heat is converted entirely into work.

An ideal monoatomic gas ($γ = 5/3$) is compressed adiabatically to $\frac{1}{8}$ of its original volume. By what factor does its pressure increase?

(A)

(B)

(C)

(D)

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✓ Correct Option: (C)

Explanation: For an adiabatic process, $P_1 V_1^\gamma = P_2 V_2^\gamma$.
$$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma = (8)^{5/3} = (8^{1/3})^5 = 2^5 = 32$$
The pressure increases by a factor of 32.

A Carnot engine operates between a high-temperature reservoir at 600 K and a low-temperature reservoir at 300 K. If the engine absorbs 800 J of heat from the hot reservoir per cycle, how much work does it perform per cycle?

(A)

400 J

(B)

200 J

(C)

600 J

(D)

500 J

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✓ Correct Option: (A)

Explanation: Carnot efficiency $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.50$ (50%).
Work done $W = \eta \times Q_H = 0.50 \times 800\text{ J} = 400\text{ J}$.

1 mole of an ideal gas expands isothermally at $T = 300\text{ K}$ from an initial volume $V$ to $2V$. Given $R = 8.314\text{ J/(mol K)}$ and $\ln(2) \approx 0.693$, what is the work done by the gas?

(A)

864 J

(B)

1728 J

(C)

2494 J

(D)

3456 J

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation: Work done in an isothermal process is $W = n R T \ln\left(\frac{V_2}{V_1}\right)$.
$$W = (1)(8.314)(300)\ln(2) = 2494.2 \times 0.693 \approx 1728\text{ J}$$

A rectangular cyclic process $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ on a $P\text{--}V$ diagram:
$•$ Path $A \rightarrow B$: Isobaric expansion at $P_2 = 300\text{ kPa}$ from $V_1 = 1m^3$ to $V_2 = 4m^3$
$•$ Path $B \rightarrow C$: Isochoric pressure drop from 300 kPa to 100 kPa at $V_2 = 4m^3$
$•$ Path $C \rightarrow D$: Isobaric compression at $P_1 = 100\text{ kPa}$ from $V_2 = 4m^3$ to $V_1 = 1m^3$
$•$ Path $D \rightarrow A$: Isochoric pressure rise from 100 kPa to 300 kPa at $V_1 = 1m^3$
Calculate the net work done by the gas in one complete cycle $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$.

(A)

200 kJ

(B)

400 kJ

(C)

300 kJ

(D)

600 kJ

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation: The net work done during a cycle on a $P\text{--}V$ diagram equals the area enclosed by the loop.
$$\text{Area} = (P_2 - P_1)(V_2 - V_1) = (300\text{ kPa} - 100\text{ kPa})(4\text{ m}^3 - 1\text{ m}^3) = 200\text{ kPa} \times 3\text{ m}^3 = 600\text{ kJ}$$

For a rigid diatomic gas (like $N_2$ at room temperature), what is the ratio of molar heat capacities $γ = \frac{C_p}{C_v}$?

(A)

1.40

(B)

1.67

(C)

1.33

(D)

1.25

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✓ Correct Option: (A)

Explanation: A rigid diatomic gas has $f = 5$ degrees of freedom (3 translational + 2 rotational).
$$C_v = \frac{5}{2}R, \quad C_p = C_v + R = \frac{7}{2}R \implies \gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = \frac{7}{5} = 1.40$$

Q10

Calculate the change in internal energy ($ΔU$) of 2 moles of a monoatomic ideal gas when its temperature increases by 40 K. ($R = 8.314\text{ J/(mol K)}$)

(A)

498.8 J

(B)

665.1 J

(C)

997.7 J

(D)

1330.2 J

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation: For a monoatomic ideal gas, internal energy is given by $U = \frac{3}{2} n R T$, so $\Delta U = \frac{3}{2} n R \Delta T$.
$$\Delta U = \frac{3}{2} \times 2 \times 8.314 \times 40 = 3 \times 8.314 \times 40 = 997.68\text{ J} \approx 997.7\text{ J}$$

Q11

A Carnot refrigerator operates between a cold reservoir at 250 K and a hot reservoir at 300 K. What is its coefficient of performance ($β$)?

(A)

4.0

(B)

5.0

(C)

6.0

(D)

0.20

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✓ Correct Option: (B)

Explanation: The coefficient of performance ($β$) for a Carnot refrigerator is:
$$\beta = \frac{T_C}{T_H - T_C} = \frac{250\text{ K}}{300\text{ K} - 250\text{ K}} = \frac{250}{50} = 5.0$$

Q12

A gas inside a closed, rigid metallic container is heated, absorbing 1500 J of thermal energy. What is the work done by the gas during this process?

(A)

1500 J

(B)

750 J

(C)

-1500 J

(D)

0 J

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation: A rigid container keeps the volume strictly constant ($ΔV = 0$). Since work done $W = \int P dV = 0$, the work done by the gas is 0 J. All added heat goes into raising internal energy ($ΔU = Q$).

Q13

Statement I: In an adiabatic expansion of an ideal gas, no heat enters or leaves the system ($Q = 0$).
Statement II: During an adiabatic expansion, the temperature of an ideal gas decreases.
Which of the following evaluations is correct regarding Statements I and II?

(A)

Both Statement I and Statement II are true, and Statement II is a valid consequence of Statement I

(B)

Both Statement I and Statement II are true, but Statement II is independent of Statement I

(C)

Statement I is true, but Statement II is false

(D)

Statement I is false, but Statement II is true

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✓ Correct Option: (A)

Explanation: In an adiabatic process $Q = 0$. By the First Law, $\Delta U = Q - W = -W$. During expansion, work is positive ($W > 0$), so $\Delta U < 0$. Since internal energy depends on temperature, a decrease in $\Delta U$ means temperature drops. Thus, Statement II directly follows from Statement I.

Q14

On a $P\text{--}V$ diagram, how does the magnitude of the slope of an adiabatic curve compare to the slope of an isothermal curve at any given point $(P, V)$?

(A)

Slope of adiabatic = Slope of isothermal

(B)

Slope of adiabatic = $\frac{1}{\gamma} \times$ (Slope of isothermal)

(C)

Slope of adiabatic = $\gamma \times$ (Slope of isothermal)

(D)

Slope of adiabatic = $\gamma^2 \times$ (Slope of isothermal)

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation: For an isothermal process ($P V = \text{const}$), $\frac{dP}{dV} = -\frac{P}{V}$.
For an adiabatic process ($P V^\gamma = \text{const}$), $\frac{dP}{dV} = -\gamma \frac{P}{V}$.
Therefore, the adiabatic slope is $\gamma$ times steeper than the isothermal slope.

Q15

1200 J of heat is transferred reversibly and isothermally to a thermal reservoir at a constant temperature of 400 K. What is the change in entropy ($ΔS$) of the reservoir?

(A)

0.33 J/K

(B)

3.00 J/K

(C)

4.00 J/K

(D)

480 J/K

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation: Change in entropy for an isothermal process is $\Delta S = \frac{Q}{T}$.
$$\Delta S = \frac{1200\text{ J}}{400\text{ K}} = 3.00\text{ J/K}$$

Q16

An ideal gas is kept in an insulated container divided into two equal compartments by a thin membrane. One side contains the gas while the other side is evacuated. The membrane ruptures, allowing the gas to freely expand into the vacuum.
What happens to the temperature of the ideal gas immediately after this free expansion process?

(A)

The temperature remains unchanged

(B)

The temperature decreases significantly

(C)

The temperature increases

(D)

The temperature drops to absolute zero

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation: During free expansion into a vacuum, no external work is done ($W = 0$) and no heat is transferred ($Q = 0$) because the container is insulated. By the First Law, $\Delta U = 0$. For an ideal gas, internal energy depends only on temperature; hence $\Delta T = 0$, and the temperature stays unchanged.

Q17

Stage State Transition Heat Input ($Q$) Work Done ($W$)
1 Expansion A $\rightarrow$ B +800 J +500 J
2 Cooling B $\rightarrow$ C -300 J 0 J
3 Compression C $\rightarrow$ D -400 J -200 J
4 Heating D $\rightarrow$ A +100 J 0 J
Refer to the thermodynamic cycle table above. What is the net work done ($W_{\text{net}}$) per cycle?

Stage	State Transition	Heat Input ($Q$)	Work Done ($W$)
1	Expansion A $\rightarrow$ B	+800 J	+500 J
2	Cooling B $\rightarrow$ C	-300 J	0 J
3	Compression C $\rightarrow$ D	-400 J	-200 J
4	Heating D $\rightarrow$ A	+100 J	0 J

(A)

+500 J

(B)

+100 J

(C)

+200 J

(D)

+300 J

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation: Net work done in a complete cycle is the sum of work done in all individual stages:
$$W_{\text{net}} = W_1 + W_2 + W_3 + W_4 = 500 + 0 + (-200) + 0 = 300\text{ J}$$
(Notice also $Q_{\text{net}} = 800 - 300 - 400 + 100 = 200...$ wait: $W_{\text{net}} = 300\text{ J}$).

Q18

A Carnot engine operating between temperatures $T_1$ and $T_2$ ($T_1 > T_2$) has an efficiency of 25%. If the sink temperature $T_2 = 300\text{ K}$, what is the source temperature $T_1$?

(A)

375 K

(B)

400 K

(C)

450 K

(D)

500 K

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation: Carnot efficiency $\eta = 1 - \frac{T_2}{T_1}$.
$$0.25 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.75 \implies T_1 = \frac{300}{0.75} = 400\text{ K}$$

Q19

For a non-linear triatomic gas molecule (such as $H_2O$ vapor, considering only translational and rotational modes), what is the molar heat capacity at constant volume ($C_v$)?

(A)

$\frac{3}{2} R$

(B)

$\frac{5}{2} R$

(C)

$3 R$

(D)

$4 R$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation: A non-linear triatomic gas has 6 degrees of freedom ($f = 6$: 3 translational + 3 rotational).
$$C_v = \frac{f}{2} R = \frac{6}{2} R = 3 R$$

Q20

A heat engine absorbs 2000 J of thermal energy from a high-temperature source and exhausts 1400 J of heat to a cold sink during each cycle. What is the thermal efficiency of this engine?

(A)

30%

(B)

70%

(C)

40%

(D)

60%

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation: Thermal efficiency $\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H}$.
$$\eta = \frac{2000 - 1400}{2000} = \frac{600}{2000} = 0.30 = 30\%$$

grade 11physicsthermodynamicscarnot enginefirst law of thermodynamicsconcept of maths

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Grade 11 : Physics - Thermodynamics (Set 4)

Grade 11 : Physics - Thermodynamics (Set 4)— Questions & Detailed Solutions

System $A$ is in thermal equilibrium with System $B$, and System $B$ is separately in thermal equilibrium with System $C$.According to the Zeroth Law of Thermodynamics, what can be concluded about Systems $A$ and $C$?

A gas expands isobarically at a constant pressure of $2.5 \times 10^5\text{ Pa}$ from an initial volume of $0.02m^3$ to a final volume of $0.06m^3$. What is the work done by the gas during this process?

During a thermodynamic process, 500 J of heat is absorbed by a system while the system performs 180 J of work on its surroundings. What is the change in internal energy ($ΔU$) of the system?

An ideal monoatomic gas ($γ = 5/3$) is compressed adiabatically to $\frac{1}{8}$ of its original volume. By what factor does its pressure increase?

A Carnot engine operates between a high-temperature reservoir at 600 K and a low-temperature reservoir at 300 K. If the engine absorbs 800 J of heat from the hot reservoir per cycle, how much work does it perform per cycle?

1 mole of an ideal gas expands isothermally at $T = 300\text{ K}$ from an initial volume $V$ to $2V$. Given $R = 8.314\text{ J/(mol K)}$ and $\ln(2) \approx 0.693$, what is the work done by the gas?

For a rigid diatomic gas (like $N_2$ at room temperature), what is the ratio of molar heat capacities $γ = \frac{C_p}{C_v}$?

Calculate the change in internal energy ($ΔU$) of 2 moles of a monoatomic ideal gas when its temperature increases by 40 K. ($R = 8.314\text{ J/(mol K)}$)

A Carnot refrigerator operates between a cold reservoir at 250 K and a hot reservoir at 300 K. What is its coefficient of performance ($β$)?

A gas inside a closed, rigid metallic container is heated, absorbing 1500 J of thermal energy. What is the work done by the gas during this process?

Statement I: In an adiabatic expansion of an ideal gas, no heat enters or leaves the system ($Q = 0$).Statement II: During an adiabatic expansion, the temperature of an ideal gas decreases.Which of the following evaluations is correct regarding Statements I and II?

On a $P\text{--}V$ diagram, how does the magnitude of the slope of an adiabatic curve compare to the slope of an isothermal curve at any given point $(P, V)$?

1200 J of heat is transferred reversibly and isothermally to a thermal reservoir at a constant temperature of 400 K. What is the change in entropy ($ΔS$) of the reservoir?

A Carnot engine operating between temperatures $T_1$ and $T_2$ ($T_1 > T_2$) has an efficiency of 25%. If the sink temperature $T_2 = 300\text{ K}$, what is the source temperature $T_1$?

For a non-linear triatomic gas molecule (such as $H_2O$ vapor, considering only translational and rotational modes), what is the molar heat capacity at constant volume ($C_v$)?

A heat engine absorbs 2000 J of thermal energy from a high-temperature source and exhausts 1400 J of heat to a cold sink during each cycle. What is the thermal efficiency of this engine?

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System $A$ is in thermal equilibrium with System $B$, and System $B$ is separately in thermal equilibrium with System $C$.
According to the Zeroth Law of Thermodynamics, what can be concluded about Systems $A$ and $C$?

Statement I: In an adiabatic expansion of an ideal gas, no heat enters or leaves the system ($Q = 0$).
Statement II: During an adiabatic expansion, the temperature of an ideal gas decreases.
Which of the following evaluations is correct regarding Statements I and II?