Heat and Thermodynamics — NEET UG practice

90 questions

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Sample questions with solutions

Q1 · 2026

An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of $75\ \text{J/s}$ , then the rate at which internal energy increases will be:

A.
75 W
B.
100 W
C.
125 W
D.
25 W

Answer: D

The first law of thermodynamics relates the heat supplied, the change in internal energy, and the work done by a system.

Q = \Delta U + W

Since the question gives rates (power) rather than total quantities, differentiate the equation with respect to time.

\frac{dQ}{dt} = \frac{d(\Delta U)}{dt} + \frac{dW}{dt}

Substitute the given rate of heat supply (100 W) and rate of work done (75 W):

100 = \frac{d(\Delta U)}{dt} + 75

Solve for the rate of increase of internal energy.

\frac{d(\Delta U)}{dt} = 25\ \text{W}

Hence, the answer is D.

Q2 · 2026

The mean free path of molecules in an ideal gas $A$ is half that of another ideal gas $B$ . The diameter of the spherical molecules of gas $A$ is twice the diameter of the molecules of $B$ . If number densities of the gases $A$ and $B$ are $n_A$ and $n_B$ , respectively, the correct option is:

A.
$n_A=\frac{1}{2} n_B$
B.
$n_A=n_B$
C.
$n_A=2 n_B$
D.
$n_A=\frac{1}{4} n_B$

Answer: A

The mean free path of gas molecules depends on molecular diameter and number density through the relation:

\lambda = \frac{1}{\sqrt{2}\pi d^2 n}

Given $\lambda_A = \frac{1}{2}\lambda_B$ and $d_A = 2d_B$ , write the mean free path expression for each gas:

\lambda_A = \frac{1}{\sqrt{2}\pi d_A^2 n_A}, \quad \lambda_B = \frac{1}{\sqrt{2}\pi d_B^2 n_B}

Dividing these and substituting the given ratio of mean free paths:

\frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A} = \frac{1}{2}

Substitute $d_A = 2d_B$ and simplify to find the ratio of number densities:

\frac{n_A}{n_B} = 2 \times \frac{d_B^2}{d_A^2} = 2 \times \frac{1}{4} = \frac{1}{2}

Hence, $n_A = \frac{1}{2}n_B$ , so the answer is A.

Q3 · 2026

One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:

A.
800 J
B.
400 J
C.
500 J
D.
600 J

Answer: D

In a full cyclic process, the gas returns to its initial state, so there is no net change in internal energy over the cycle.

\Delta U = 0

From the first law of thermodynamics, the total heat supplied equals the total work done by the gas over the cycle.

\Delta Q = W + \Delta U = W

From the $P$ - $V$ diagram, the work done in one cycle equals the enclosed area, which for this rectangular cycle is the product of the pressure difference and the volume difference.

W = (300-100)(5-2) = 200 \times 3 = 600\ \text{J}

Hence, the total heat supplied is 600 J, so the answer is D.

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