Question

Under the isothermal condition, a gas at $300\mathrm{K}$ expands from $0.1\mathrm{L}$ to $0.25\mathrm{L}$ against a constant external pressure of $2\mathrm{bar}$. Find the work done by the gas. It is given that $1\mathrm{L}‐\mathrm{bar}=100\mathrm{J}$.

• A. $5\text{kJ}$
• B. $-30\text{J}$
• C. $30\text{J}$
• D. $25\text{J}$

Answer 1

The correct option is B

$-30\text{J}$

Step 1: Given parameters

The temperature of the gas, $T=300\text{K}$

Initial volume, $V_1=0.1\text{L}$

Final volume, $V_2=0.25\text{L}$

External pressure, $P=2\text{bar}$

Step 2: To find

We have to find the work done by the gas.

Step 3: Calculating work done

A gas expanding against a constant external pressure is an irreversible process. The work is done in an irreversible process.

Now,

$$\begin{gathered} W=-P∆V \\ W=-P\left(V_2-V_1\right) \end{gathered}$$

Here,

W is the work done.

P is the pressure.

V is the volume.

The negative ($-$) sign indicates the system loses energy.

By substituting the given values in the above formula, we get

$$\begin{gathered} \mathrm{W}=-2\mathrm{bar}\left(0.25\mathrm{L}-0.1\mathrm{L}\right) \\ =-2\times 0.15\mathrm{L}\mathrm{bar} \\ =-0.30\mathrm{L}\mathrm{bar}(1\mathrm{L}\mathrm{Bar}=100\mathrm{J}) \\ =-0.30\times 100\mathrm{J} \\ =-30\mathrm{J} \end{gathered}$$

Hence, option (B) is correct.