Question

A gas expands by $0.5$ L against a constant pressure of $1$ atm. The work done in calories is:

• A. $-50.66$
• B. $-12.10$
• C. $-5.64$
• D. $-35.61$

Answer 1

Answer: (B)

$-12.10$

The expression for the work done is given below:

$w=-P\Delta V$, where $P$ is the pressure and $\Delta V$ is the change in volume.

Substituting the values, we get

$w=-1\times 0.5$ $=-0.5$ atm L

$w=-0.5\times \frac{101.325}{4.18}$ cal $=-12.1$ cal