Question

During isothermal expansion of one mole of an ideal gas from 10 atm to 1 atm at 273 K, the work done is:

• A. – 895.8 cal
• B. – 1172.6 cal
• C. – 1257.43 cal
• D. – 1499.6 cal

Answer 1

The correct option is C – 1257.43 cal

Work done in a reversible isothermal process of gas (w) = $-2.303\times nRTlog\frac{P_1}{P_2}$,
where, n is the number of moles,
R is the gas constant,
T is temperature and
$P_1$ and $P_2$ are the initial and final pressure of the gas.

According to the given conditions,
w = $-2.303\times 1\times 2\times 273log\frac{10}{1}$
= –1257.43 cal.