Question

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

• A. $-10368J$
• B. $-110368J$
• C. $12000J$
• D. $120368J$

Answer 1

The correct option is A $-10368J$

Work done in an isothermal process is given by
$W=2.3026nRT{\log }_{10}\frac{V_2}{V_1}$
Here, $n=3,R=8.31J/mo{l}^{o}C$
$T=300K$, $V_1=4L$, $V_2=1L$
Hence, $W=2.3026\times 3\times 8.31\times 300\times lo{g}_{10}\frac{1}{4}$
= $17221.15(-2{\log }_{10}2$)
= $-17221.15\times 2\times 0.3010=-10368J$