Question

A monoatomic gas is supplied heat 2 Q very slowly keeping the pressure, constant. The work done by the gas is

• A. $\frac{4}{5}$Q
• B. $\frac{3}{5}$Q
• C. $\frac{7}{5}$Q
• D. Q

Answer 1

Answer: (A)

$\frac{4}{5}$Q

Answer 2

For a constant pressure process, the heat supplied is given by

$nC_p\Delta T = 2Q$.

For a monoatomic gas,

$C_p = \frac{5}{2} R$.

The work done is

$W = nR\Delta T$.

Express $\Delta T$ from the heat equation:

$\Delta T = \frac{2Q}{nC_p} = \frac{2Q}{n\frac{5}{2}R} = \frac{4Q}{5nR}$.

Thus,

$W = nR \times \frac{4Q}{5nR} = \frac{4}{5}Q$.