Question

Three moles of an ideal gas kept at a constant temperature at $300\mathrm{k}$ are compressed from a volume of $4\mathrm{L}$ to $1\mathrm{L}$. The work done in the process is

• A. $-10368\mathrm{J}$
• B. $-110368\mathrm{J}$
• C. $12000\mathrm{J}$
• D. $120368\mathrm{J}$

Answer 1

The correct option is A

$-10368\mathrm{J}$

Step 1: Given parameters

Numbers of moles of an ideal gas $\mathrm{n}=3$

Temperature, $\mathrm{T}=300\mathrm{K}$

Volume, $$\begin{gathered} {\mathrm{V}}_1=4\mathrm{litre} \\ {\mathrm{V}}_2=1\mathrm{litre} \end{gathered}$$

Step 2: To find

We have to find the work done.

Step 3: Calculation

We know that in an isothermal process,

The work done:

$\mathrm{W}=2.3026{\mathrm{nRTlog}}_{10}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

By substituting the values,

$$\begin{gathered} =2.3026\times 3\times 8.31\times 300\times {\log }_{10}\frac{1}{4} \\ =-10368\mathrm{J} \end{gathered}$$

Therefore, the work done in the process is $-10368\mathrm{J}$.