Question

Quantity of work (in joules) done by the gas if it expands against a constant pressure of $0.980atm$ and the change in volume $(\Delta V)$ is $25.0L,$ is

• A. 24.5 J
• B. 2.48 J
• C. $2.48\times 10^{3}J$
• D. 0.0245 J

Answer 1

Answer: (C)

Work done $=p\Delta V$

$=0.980atm\times 25.0L=24.5Latm$

$=\frac{24.5Latm\times 8.314Jmo{l}^{-1}{K}^{-1}}{0.0821Latmmo{l}^{-1}{K}^{-1}}=2481J=2.481\times 10^{3}J$

Note Use value of $R$ (gas constant to convert $L$ atm to $J$

$R=0.0821L$ atm $mo{l}^{-1}{K}^{-1}=8.314Jmo{l}^{-1}{K}^{-1}$