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

The Correct Option is A

For an isobaric process:
The work done is given by:
\[ w = P \Delta v = nR \Delta T = 100 \, \text{J} \]

The first law of thermodynamics states:
\[ Q = \Delta U + W \]

For an ideal gas, the change in internal energy is:
\[ \Delta U = \frac{f}{2} nR \Delta T \] Therefore,
\[ \Delta Q = \frac{f}{2} nR \Delta T + nR \Delta T \] \[ \Delta Q = \left( \frac{f}{2} + 1 \right) nR \Delta T \]

Substituting \( f = 5 \) and \( nR \Delta T = 100 \, \text{J} \):
\[ \Delta Q = \left( \frac{5}{2} + 1 \right) 100 = 350 \, \text{J} \]

Final Answer:
\[ Q = 350 \, \text{J} \]

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:
\[Q = 350 \, \text{J}.\]