Question

During isothermal expansion of one mole of an ideal gas from $10atm$ to $1atm$ at $273K$, the work done is: [Gas constant = $2$]

• A. $-895.8cal$
• B. $-1172.6cal$
• C. $-1257.43cal$
• D. $-1499.6cal$

Answer 1

Answer: (C)

$-1257.43cal$

Work done in expansion of gas during isothermal process,

$=-2.303nRT\log \frac{P_1}{P_2}$

$=-2.303\times 1\times 2\times 273\log \frac{10}{1}$

$=-1257.42cal$

Hence, option C is correct