Question

A gas is compressed in a cylinder from a volume of 20.0 l to 2.0 l by a constant pressure of 10.0 atm. calculate the amount of work done on the system.

Answer 1

First, we need to convert the pressure in SI units. Keeping in mind that $1 atm = 1.01 \cdot 10^5 Pa$:
$p=10.0 atm =1.01 \cdot 10^6 Pa$

The initial and final volume of the gas are (keeping in mind that $1.0 L = 0.001 m^3$):
$V_i = 20.0 L=0.020 m^3$
$V_f = 2.0 L=0.002 m^3$

So, the work done on the gas by the surrounding is
$W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J$
And the final positive sign means that this work corresponds to an increase in internal energy of the gas.