Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:

Options

A. ` 572 J`
B. `735 J `
C. ` 441 J`
D. `525 J`

Answer

To solve the problem, we need to determine the increase in the internal energy of a diatomic gas when it is given 735 J of heat energy at constant pressure.

### Step-by-Step Solution:

1. **Understand the Process**: The gas is expanding at constant pressure (isobaric process). In an isobaric process, the heat added to the system (Q) is related to the change in internal energy (ΔU) and the work done (W) by the gas.

2. **Use the First Law of Thermodynamics**: The first law states that:
\[ Q = \Delta U + W \]
Rearranging gives us:
\[ \Delta U = Q - W \]

3. **Calculate Work Done (W)**: For an isobaric process, the work done by the gas when it expands is given by:
\[ W = P \Delta V \]
However, we can also express it in terms of the number of moles (n) and the change in temperature (ΔT):
\[ W = nR\Delta T \]

4. **Relate Heat to Temperature Change**: The heat added at constant pressure is related to the change in temperature by:
\[ Q = nC_p \Delta T \]
For a diatomic gas, the molar heat capacity at constant pressure \(C_p\) is:
\[ C_p = \frac{7}{2}R \]
Therefore, we can write:
\[ Q = n \left(\frac{7}{2}R\right) \Delta T \]

5. **Express ΔU in Terms of Q**: The change in internal energy for a diatomic gas is given by:
\[ \Delta U = nC_v \Delta T \]
where \(C_v\) for a diatomic gas is:
\[ C_v = \frac{5}{2}R \]
Thus:
\[ \Delta U = n \left(\frac{5}{2}R\right) \Delta T \]

6. **Substituting ΔT**: From the equation for Q, we can express ΔT as:
\[ \Delta T = \frac{2Q}{7nR} \]
Substituting this into the equation for ΔU gives:
\[ \Delta U = n \left(\frac{5}{2}R\right) \left(\frac{2Q}{7nR}\right) \]
Simplifying this yields:
\[ \Delta U = \frac{5Q}{7} \]

7. **Calculate ΔU**: Now, substituting the value of Q (735 J):
\[ \Delta U = \frac{5 \times 735}{7} \]
\[ \Delta U = \frac{3675}{7} = 525 \text{ J} \]

### Final Answer:
The increase in the internal energy of the gas is **525 J**.

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