Question

A diatomic gas ($\gamma = 1.4$) does 100 J of work in an isobaric expansion. The heat given to the gas is:

• A. 350 J
• B. 490 J
• C. 150 J
• D. 250 J

Answer 1

Answer: (A)

350 J

Answer 2

For an isobaric process, the work done is:
$w = P \Delta V = nR \Delta T.$
Given:
$w = 100 \, \text{J}.$
The heat ($Q$) supplied is:
$Q = \Delta U + w,$
where the internal energy change ($\Delta U$) for a diatomic gas is:
$\Delta U = \frac{f}{2} nR \Delta T,$
with $f = 5$ (degrees of freedom for a diatomic gas).
Substitute :
$Q = \frac{f}{2} nR \Delta T + nR \Delta T.$
Simplify :
$Q = \left(\frac{f}{2} + 1\right) nR \Delta T.$
Substitute $\Delta T$ from $w = nR \Delta T$:
$Q = \left(\frac{f}{2} + 1\right) \times w.$
Substitute $f = 5$ and $w = 100 \, \text{J}$:
$Q = \left(\frac{5}{2} + 1\right) \times 100 = \left(\frac{7}{2}\right) \times 100 = 350 \, \text{J}.$
Thus, the heat given to the gas is:
$Q = 350 \, \text{J}.$