MCQEasyJEE 2025First Law & Internal Energy

JEE Physics 2025 Question with Solution

An ideal gas goes from an initial state to final state. During the process, the pressure of the gas increases linearly with temperature.

A. The work done by gas during the process is zero.

B. The heat added to the gas is different from the change in its internal energy.

C. The volume of the gas is increased.

D. The internal energy of the gas is increased.

E. The process is isochoric (constant volume process).

Choose the correct answer from the options given below:

  • A

    A, B, C, D Only

  • B

    A, D, E Only

  • C

    E Only

  • D

    A, C Only

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The pressure of an ideal gas increases linearly with temperature.

Find: Which statements among A, B, C, D, E are correct.

For an ideal gas,

PV=nRTPV = nRT

If PP increases linearly with TT, then PT=nRV\frac{P}{T} = \frac{nR}{V} remains constant, so VV remains constant. Therefore, the process is isochoric.

Now examine each statement:

  1. A. In an isochoric process, ΔV=0\Delta V = 0, so work done by the gas is
W=PdV=0W = \int P \, dV = 0

Hence, A is true.

  1. B. From the first law of thermodynamics,
Q=ΔU+WQ = \Delta U + W

Since W=0W = 0 for an isochoric process,

Q=ΔUQ = \Delta U

So the heat added is not different from the change in internal energy. Hence, B is false.

  1. C. Since volume remains constant, the volume does not increase. Hence, C is false.

  2. D. For an ideal gas, internal energy depends only on temperature. As temperature increases, internal energy increases. Hence, D is true.

  3. E. Because PTP \propto T for constant VV, the process is isochoric. Hence, E is true.

Therefore, the correct statements are A, D, E only. The correct option is B.

Statement-wise Analysis

Given: Pressure increases linearly with temperature for an ideal gas.

Find: The correct combination of statements.

Using the ideal gas equation,

P=nRVTP = \frac{nR}{V}T

This is a linear relation between PP and TT when nn and VV are constant. Hence the process is at constant volume.

Because the process is isochoric:

ΔV=0\Delta V = 0

so,

W=PdV=0W = \int P \, dV = 0

Also, from the first law,

Q=ΔU+W=ΔUQ = \Delta U + W = \Delta U

Thus heat supplied equals the change in internal energy, not different from it.

Further, for an ideal gas,

UTU \propto T

So when temperature increases, internal energy increases. Since volume is constant, there is no increase in volume.

Hence:

  • A true
  • B false
  • C false
  • D true
  • E true

Therefore, the correct option is B.

Common mistakes

  • Assuming that pressure increasing with temperature automatically means volume also increases. For an ideal gas, PTP \propto T at constant volume, so this relation actually indicates an isochoric process. Use PV=nRTPV = nRT before concluding how volume changes.

  • Using W=PΔVW = P\Delta V without checking whether volume changes. In an isochoric process, ΔV=0\Delta V = 0, so the work done is zero. First identify the process, then evaluate work.

  • Thinking that heat added must always be different from the change in internal energy. By the first law, Q=ΔU+WQ = \Delta U + W. When W=0W = 0, heat added equals the change in internal energy. Always account for the work term.

Practice more First Law & Internal Energy questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions