Question

A mono atomic gas is supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

Step 1: Given that:

The heat supplied to the monoatomic gas= $Q$

Heat is supplied at constant pressure.

Step 2: Calculation of work done:

According to first law of thermodynamics

$Q=\Delta U+W$

Where $\Delta U$ is the change in internal energy and $W$ is the amount of work done.

Thus, $W=Q-\Delta U$

The internal energy of a gas is given by;

$\Delta U=n{C}_V\Delta T$

Where n is the number of moles of the gas and $C_V$ is the specific heat of the gas at constant volume.

Heat supplied to the gas is given by;

$Q=n{C}_P\Delta T$

Where $C_P$ is the specific heat of the gas at constant pressure.

Now,

$\frac{\Delta U}{Q}=\frac{n{C}_V\Delta T}{n{C}_P\Delta T}$

$\frac{\Delta U}{Q}=\frac{C_V}{C_P}$

The ratio of specific heat of a gas at constant pressure and constant volume is given as;

$\frac{C_P}{C_V}=\text{γ}=1+\frac{2}{f}$

Where f is the degree of freedom.

For monoatomic gas, $f=3$

Thus,

$\frac{C_P}{C_V}=1+\frac{2}{3}$

$\Rightarrow \frac{C_P}{C_V}=\frac{5}{3}$

$\Rightarrow \frac{C_V}{C_P}=\frac{3}{5}$

Thus,

$\frac{\Delta U}{Q}=\frac{3}{5}$

$\Rightarrow \Delta U=\frac{3}{5}Q$

Hence

The work done will be

$W=Q-\frac{3}{5}Q$

$W=\frac{2}{5}Q$

Thus,

Option C) $\frac{2}{5}Q$ is the correct option.