Q: During an isobaric expansion of 2\text{ moles} of an ideal gas, the temperature rises by 50\text{ K}. The work done by the gas is (R = 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}):

831.4\text{ J}. At constant pressure W = P\,\Delta V = nR\,\Delta T = 2 \times 8.314 \times 50 = 831.4\text{ J}.

Question 1

The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system, then:

Accepted Answer

They are in thermal equilibrium with each other. This law establishes temperature as a valid physical property: bodies in mutual thermal equilibrium share a common temperature, which is what a thermometer measures.

Question 2

Which of the following is NOT a thermodynamic state variable?

Accepted Answer

Work done. Pressure, volume, temperature and internal energy depend only on the current state of the system. Work and heat depend on the path taken, so they are path functions, not state variables.

Question 3

The work done by a gas expanding at constant pressure P from volume V_1 to V_2 is:

Accepted Answer

P(V_2 - V_1). For an isobaric process W = \int P\,dV = P\int dV = P\,\Delta V = P(V_2 - V_1).

Question 4

During an isobaric expansion of 2	ext{ moles} of an ideal gas, the temperature rises by 50	ext{ K}. The work done by the gas is (R = 8.314	ext{ J}\cdot	ext{mol}^{-1}\cdot	ext{K}^{-1}):

Accepted Answer

831.4	ext{ J}. At constant pressure W = P\,\Delta V = nR\,\Delta T = 2 	imes 8.314 	imes 50 = 831.4	ext{ J}.

Question 5

On a P-V indicator diagram, the work done by a gas during a process is represented by:

Accepted Answer

The area under the curve. Since W = \int P\,dV, the work is the area between the process curve and the volume axis.

Question 6

For a gas taken around a complete cyclic process, the change in internal energy is:

Accepted Answer

Zero. Internal energy is a state function, so returning to the initial state means \Delta U = 0, whatever path the cycle follows.

Question 7

In a cyclic process, the net heat absorbed by the system is equal to:

Accepted Answer

The net work done by the system. The first law gives Q = \Delta U + W, and with \Delta U = 0 over a cycle, Q = W.

Question 8

The net work done by a gas during a cyclic process on a P-V diagram is equal to:

Accepted Answer

The area enclosed by the closed loop. Expansion contributes positive area and compression negative area, so the net work is the area enclosed by the cycle.

Question 9

A cyclic process traced in the clockwise sense on a P-V diagram represents a process in which:

Accepted Answer

Net work is done by the gas and heat is converted into work. A clockwise loop expands at high pressure and compresses at low pressure, so the positive area dominates. This is the cycle of a heat engine.

Question 10

In an isochoric (constant volume) process, the work done by the gas is:

Accepted Answer

Zero. With \Delta V = 0, W = P\,\Delta V = 0. All the heat supplied therefore goes into raising the internal energy.

Question 11

A quasi-static thermodynamic process is one that is carried out:

Accepted Answer

So slowly that the system remains in equilibrium at every stage. Only if the process is infinitesimally slow can the pressure and temperature be well defined throughout, which is what allows the state to be plotted as a curve.

Question 12

Which of the following pairs correctly classifies the quantities as intensive and extensive respectively?

Accepted Answer

Temperature and volume. Intensive properties (temperature, pressure, density) do not depend on the amount of matter; extensive properties (volume, mass, internal energy) scale with it.

Question 13

Internal energy is a state function whereas heat and work are:

Accepted Answer

Path functions, depending on how the change is brought about. Two different paths between the same pair of states give the same \Delta U but generally different Q and W.

Question 14

Two bodies are said to be in thermal equilibrium when:

Accepted Answer

There is no net flow of heat between them. Thermal equilibrium means both are at the same temperature, so heat flows equally in both directions and there is no net transfer.

Question 15

A wall that permits no exchange of heat between a system and its surroundings is called:

Accepted Answer

An adiabatic wall. An adiabatic (perfectly insulating) wall blocks heat flow, whereas a diathermic wall conducts heat freely and allows the two sides to reach a common temperature.

Question 16

By the usual sign convention in thermodynamics, work done by the system and heat absorbed by the system are taken respectively as:

Accepted Answer

Positive and positive. Heat added to the system and work done by the system are both counted positive, which is why the first law is written Q = \Delta U + W.

Question 17

A gas expands by 0.01	ext{ m}^3 against a constant external pressure of 2 	imes 10^5	ext{ Pa}. The work done by the gas is:

Accepted Answer

2000	ext{ J}. W = P\Delta V = 2 	imes 10^5 	imes 0.01 = 2000	ext{ J}.

Question 18

Three moles of an ideal gas are heated at constant pressure so that the temperature rises by 40	ext{ K}. The work done by the gas is (R = 8.314):

Accepted Answer

998	ext{ J}. W = nR\Delta T = 3 	imes 8.314 	imes 40 \approx 998	ext{ J}.

Question 19

A gas at a constant pressure of 10^5	ext{ Pa} expands from 0.02	ext{ m}^3 to 0.05	ext{ m}^3. The work done is:

Accepted Answer

3000	ext{ J}. W = P(V_2 - V_1) = 10^5 	imes (0.05 - 0.02) = 10^5 	imes 0.03 = 3000	ext{ J}.

Question 20

A gas is carried around a clockwise cycle enclosing an area of 500	ext{ J} on a P-V diagram. The net heat absorbed per cycle is:

Accepted Answer

500	ext{ J}. Over a cycle \Delta U = 0, so Q = W. The clockwise area gives W = +500	ext{ J}, hence Q = 500	ext{ J}.

Thermodynamic Systems & Zeroth Law — MH-CET Physics MCQs with Solutions

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