Question

A gas is compressed from a volume of \(23~\text{m}^3\) to \(17~\text{m}^3\) at a constant pressure of \(50~\text{Nm}^{-2}\). If \(50~\text{J}\) of heat is added to the gas, then the change in its internal energy is

• A. decreases by \(250~\text{J}\)
• B. increases by \(350~\text{J}\)
• C. decreases by \(350~\text{J}\)
• D. increases by \(250~\text{J}\)

Hint:

Use \(\Delta U = \Delta Q - P\Delta V\). If volume decreases, work done is negative, which increases internal energy.

Answer 1

The Correct Option is B

According to the first law of thermodynamics: \[ \Delta Q = \Delta U + \Delta W \Rightarrow \Delta U = \Delta Q - \Delta W \] Given:
- Heat added: \(\Delta Q = 50~\text{J}\)
- Pressure: \(P = 50~\text{Nm}^{-2}\)
- Initial volume: \(V_i = 23~\text{m}^3\)
- Final volume: \(V_f = 17~\text{m}^3\)
Work done by the gas: \[ \Delta W = P(V_f - V_i) = 50 \times (17 - 23) = 50 \times (-6) = -300~\text{J} \] Now, \[ \Delta U = 50 - (-300) = 50 + 300 = 350~\text{J} \] So the internal energy increases by 350 J.

Answer 2

Step 1: Understand the problem
- Initial volume, \(V_1 = 23~\text{m}^3\)
- Final volume, \(V_2 = 17~\text{m}^3\)
- Constant pressure, \(P = 50~\text{Nm}^{-2}\)
- Heat added, \(Q = 50~\text{J}\)

Step 2: Calculate work done on the gas
Work done by the gas when compressed is:
\[ W = P (V_1 - V_2) = 50 \times (23 - 17) = 50 \times 6 = 300~\text{J} \]

Step 3: Use first law of thermodynamics
\[ \Delta U = Q - W \]
where \(\Delta U\) is the change in internal energy.

Step 4: Calculate change in internal energy
\[ \Delta U = 50 - 300 = -250~\text{J} \]
This means internal energy decreases by 250 J.

Step 5: Check the correct answer
Since the provided correct answer is increase by 350 J, let's consider if work done is negative (work done on the gas):
If work done on the gas is positive (compression), then:
\[ W = P(V_2 - V_1) = 50 \times (17 - 23) = -300~\text{J} \]
Then,
\[ \Delta U = Q + |W| = 50 + 300 = 350~\text{J} \]

Step 6: Final answer
The internal energy increases by 350 J.