Question

Suppose a mixture of gases occupies 5 L. Combustion occurs that causes the gases to expand to 25 L. If the pressure kept a constant 3 atm, what is the work done on the system?

Answer 1

The work done by a gas at contant pressure is given by:

$\begin{gathered} W=F\times d \\ P=\frac{F}{A}\therefore F=P\times A \\ \\ W=P\times A\times d\text{ }But,\text{ }A\times d=V \\ W=P\Delta V \\ P:pressure(Pa)1atm=101325Pa \\ F:force \\ A:area \\ V:volume(m^3)=\frac{L}{1000}=m^3 \\ d:distance \\ W:workdone \\ \Delta V:(V_f-V_i) \\ \end{gathered}$

By substituting the known values in the equation we have:

$\begin{gathered} W=303975\text{ }Nm^{-2}\times(0.025m^3-0.005m^3) \\ W=6079.5Nm \\ \\ W=6,079.5J \end{gathered}$

Answer: Work done is +6,079.5J.