Question

A gas is compressed from a volume of $2 \,m^3$ to a volume of $1\, m^3$ at a constant pressure of $100\, N/m^2$. Then it is heated at constant volume by supplying $150\, J$ of energy. As a result, the internal energy of the gas :

• A. Increases by 250 J
• B. Decreases by 250 J
• C. Increases by 50 J
• D. Decreases by 50 J

Answer 1

The Correct Option is A

To solve this problem, we'll use the First Law of Thermodynamics, which is given by:

\Delta U = Q - W

where:

• \Delta U is the change in internal energy.
• Q is the heat supplied to the system.
• W is the work done by the system.

Let's analyze each step of the process given in the question:

1. **Compression Step:** The gas is compressed from 2 \, m^3 to 1 \, m^3 at a constant pressure of 100 \, N/m^2.

The work done on the gas during this constant pressure compression is calculated as:

W = P \cdot \Delta V = P \cdot (V_{\text{final}} - V_{\text{initial}})

Substitute the given values:

W = 100 \, N/m^2 \times (1 \, m^3 - 2 \, m^3) = 100 \times (-1) = -100 \, J

Here, the work done is negative because the volume decreases, meaning work is done on the gas.

2. **Heating Step:** The gas is heated at constant volume by supplying 150 \, J of energy.

Thus, the heat added to the gas Q = 150 \, J.

Substitute Q and W into the First Law of Thermodynamics:

\Delta U = Q - W = 150 \, J - (-100 \, J)

\Delta U = 150 \, J + 100 \, J = 250 \, J

Therefore, the internal energy of the gas increases by 250 \, J.

Conclusion: The correct answer is that the internal energy of the gas increases by 250 J.