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MCQEasyJEE 2023Isothermal & Adiabatic Processes

JEE Physics 2023 Question with Solution

Heat is given to an ideal gas in an isothermal process.

A. Internal energy of the gas will decrease.

B. Internal energy of the gas will increase.

C. Internal energy of the gas will not change.

D. The gas will do positive work.

E. The gas will do negative work.

Choose the correct answer from the options given below:

  • A

    A and E only

  • B

    B and D only

  • C

    C and E only

  • D

    C and D only

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Heat is given to an ideal gas in an isothermal process.

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

In an isothermal process, temperature remains constant.

dT=0dT = 0

For an ideal gas, internal energy depends only on temperature. Therefore,

dU=0dU = 0

So the internal energy does not change. Hence, C is correct, while A and B are incorrect.

From the first law of thermodynamics,

dQ=dU+dWdQ = dU + dW

Since

dU=0dU = 0

we get

dQ=dWdQ = dW

Heat is given to the gas, so

dQ>0dQ > 0

Therefore,

dW>0dW > 0

So the gas does positive work. Hence, D is correct and E is incorrect.

The solution contains a contradictory line stating "The correct answer is (C): (c), (e)." However, the working clearly shows that C and D are correct.

Therefore, the correct option is D.

Using internal energy and first law

Given: The process is isothermal and heat is supplied to an ideal gas.

Find: The correct combination of statements.

  1. In an isothermal process,
dT=0dT = 0
  1. For an ideal gas, internal energy is a function only of temperature. Hence,
dU∝dTdU \propto dT

and therefore,

dU=0dU = 0

Thus statement C is true.

  1. Apply the first law:
dQ=dU+dWdQ = dU + dW

Substituting

dU=0dU = 0

we obtain

dQ=dWdQ = dW
  1. Since heat is given to the gas,
dQ>0dQ > 0

so,

dW>0dW > 0

Thus statement D is true.

  1. Therefore statements A, B, and E are false.

Hence the correct combination is C and D only, which corresponds to option D.

Common mistakes

  • Assuming that giving heat to a gas must increase internal energy is incorrect here because for an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature remains constant, so use dU=0dU = 0.

  • Confusing the sign of work is a common error. When heat is supplied in an isothermal expansion, the gas does work on the surroundings, so the work done by the gas is positive, not negative.

  • Using the first law without the isothermal condition leads to mistakes. First apply dT=0dT = 0, then infer dU=0dU = 0, and only after that reduce dQ=dU+dWdQ = dU + dW to dQ=dWdQ = dW.

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