Question

Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 18.0 atm. Calculate the work, w , if the gas expands against a constant external pressure of 1.00 atm to a final volume of 18.0 L.

Answer 1

Answer:

-32821.2 J (negative sign implies that work is done by the system)

Explanation:

The expression for the calculation of work done is shown below as:

$w=-P\times \Delta V$

Where, P is the pressure

$\Delta V$ is the change in volume

From the question,  

$\Delta V$ = 18.0 - 1.00 L = 18.0 L

P = 18.0 atm

$w=-18.0\times18.0\ atmL$

Also, 1 atmL = 101.3 J

So,  

$w=-18.0\times 18.0\times 101.3\ J=-32821.2\ J$ (negative sign implies that work is done by the system)