Question

A gas is compressed at a constant pressure of \(50\,N/m^2\) from a volume of \(10\,m^3\) to a volume of \(4\,m^3\). Energy of \(100\,J\) is then added to the gas by heating. Its internal energy is

• A. Increases by \(400\,J\)
• B. Increases by \(200\,J\)
• C. Increases by \(100\,J\)
• D. Decreases by \(200\,J\)

Hint:

During compression, work done by gas is negative, so internal energy may increase more than the heat supplied.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem applies the First Law of Thermodynamics, which relates the change in internal energy ($\Delta U$) of a system to the heat added to the system ($Q$) and the work done by the system ($W$).
Step 2: Key Formula or Approach:
1. The First Law of Thermodynamics is given by: $\Delta U = Q - W$. 2. Work done by the gas during a constant pressure process (isobaric) is $W = P \Delta V = P(V_{final} - V_{initial})$. 3. We need to be careful with the sign convention. $Q$ is positive if heat is added to the system. $W$ is positive if work is done by the system (expansion), and negative if work is done on the system (compression).
Step 3: Detailed Explanation:
Given: - Constant pressure, $P = 50$ N/m$^2$. - Initial volume, $V_{initial} = 10$ m$^3$. - Final volume, $V_{final} = 4$ m$^3$. - Heat added to the gas, $Q = +100$ J. First, calculate the work done by the gas: \[ W = P(V_{final} - V_{initial}) \] \[ W = 50 \times (4 - 10) \] \[ W = 50 \times (-6) = -300 \text{ J} \] The work done by the gas is negative, which makes sense as the gas is compressed (work is done {on} the gas). Now, use the First Law of Thermodynamics to find the change in internal energy, $\Delta U$: \[ \Delta U = Q - W \] \[ \Delta U = (+100) - (-300) \] \[ \Delta U = 100 + 300 = 400 \text{ J} \] The change in internal energy is $+400$ J, which means the internal energy increases by 400 J. Step 4: Final Answer:
The internal energy of the gas increases by 400 J. Therefore, option (A) is correct.