Question:

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:

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In processes involving ideal gases, when the volume remains constant, the work done is zero. The heat added in such a case is equal to the change in internal energy. This type of process is called an isochoric process.
Updated On: Nov 14, 2025
  • A, B, C, D Only
  • A, D, E Only
  • E Only
  • A, C Only
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The Correct Option is B

Approach Solution - 1

Let's analyze the given processes individually: 

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

In thermodynamics, work done by a gas is zero in an isochoric process, where volume remains constant. Since the question suggests the pressure increases linearly with temperature, the gas follows an isochoric process if option E is correct. Thus, the work done is indeed zero under these conditions.

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

For an ideal gas, the change in internal energy (\(\(\Delta U\)\)) is directly proportional to the change in temperature (\(\(\Delta T\)\)). According to the first law of thermodynamics:

\(\Delta Q = \Delta U + \Delta W\)

Since option E suggests an isochoric process, \(\Delta W = 0\). Therefore, \(\Delta Q = \Delta U\). This option is false under an isochoric process.

  • Option C: The volume of the gas is increased.

If pressure increases linearly with temperature and volume is constant, the volume does not increase. Thus, this option is also false.

  • Option D: The internal energy of the gas is increased.

The internal energy of an ideal gas is a function of temperature. Since temperature increases (as pressure increases linearly with it), the internal energy increases. Thus, option D is true.

  • Option E: The process is isochoric (constant volume process).

If pressure increases linearly with temperature, it means volume remains constant as per the ideal gas law \(PV = nRT\). A linear increase of pressure with temperature typically defines an isochoric process where volume does not change.

After evaluating all options, only A, D, and E are correct. Hence, the correct answer is:

A, D, E Only

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Approach Solution -2

Given: We are told that the pressure of the gas increases linearly with temperature. The relationship between pressure and temperature for an ideal gas is given by:

\[ P = aT + b, \] where \( a \) and \( b \) are constants. This suggests that the pressure increases linearly with temperature.

Step 1: Analyzing the options

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

Since the pressure increases linearly with temperature, and for an ideal gas undergoing this process, the volume remains constant, there is no change in volume. The work done by the gas is given by: \[ W = P \Delta V. \] Since \( \Delta V = 0 \), the work done by the gas is zero. Thus, Option A is correct.

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

For an ideal gas undergoing any process, the heat added is related to both the work done and the change in internal energy. Since there is no change in volume in this case, the process is isochoric (constant volume), meaning all the heat added goes into increasing the internal energy. Therefore, Option B is incorrect because the heat added is equal to the change in internal energy.

Option C: The volume of the gas is increased.

As we established earlier, since this is an isochoric process (constant volume), the volume of the gas does not change. Thus, Option C is incorrect.

Option D: The internal energy of the gas is increased.

For an ideal gas, the internal energy depends only on the temperature. Since the temperature increases in this process, the internal energy of the gas increases. Thus, Option D is correct.

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

Given that no work is done by the gas and the pressure increases with temperature, this suggests that the volume remains constant. Hence, Option E is correct.

Final Answer:

The correct answer is \( \boxed{A, D, E \text{ Only}} \).

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