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* d \\ P=(F)/(A)\therefore F=P* A \\ \\ W=P* A* d\text{ }But,\text{ }A* d=V \\ W=P\Delta V \\ P:pressure(Pa)1atm=101325Pa \\ F:force \\ A:area \\ V:volume(m^3)=(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)*(0.025m^3-0.005m^3) \\ W=6079.5Nm \\ \\ W=6,079.5J \end{gathered}`

Answer: Work done is +6,079.5J.