Question

Calculate the work done during isothermal expansion of 1 mol of an ideal gas from 10 atm to 1 atm at 300 K.

• A. - 1382 cal
• B. - 1382 J
• C. - 13.82 cal
• D. - 1.382 cal

Answer 1

The correct option is A - 1382 cal

Given;
T= 300K
$P_1=10$
$P_2=1$
$n=1$
$R=8.314$
$w=?$

Formula
$w=-2.303nRTlog\frac{p_1}{p_2}$

$=-2.303\times 8.314\times 300$
$=-5744.14Jor1374.19cal$