Question

A sample of gas occupies a volume of 70.9 mL. As it expands, it does 118.9 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?

Answer 1

`V_2=1.07L`

Explanation:

The relationship between work (`w`), pressure (`P`) and volume (`V`) is the following:

`w=-PDeltaV`

where, `DeltaV=V_2-V_1`

since the gas is expanding, then the work is done by the system and it is of a negative value .

Note that work, in this case, should be expressed in `L*atm`.

`1L*atm=101.3J` therefore,
`w=118.9cancel(J)xx(1L*atm)/(101.3cancel(J))=1.174L*atm`

Since work is done by the system: `w=-1.174L*atm`

Pressure should then be expressed in `atm`:

`P=783cancel("torr")xx(1atm)/(760cancel("torr"))=1.03atm`

Thus, replacing every term in its value in the expression `w=-PDeltaV` we get:

`cancel(-)1.174L*cancel(atm)=cancel(-)1.03cancel(atm)xxDeltaV`

`=>DeltaV=(1.174)/(1.03)=1.14L`

Note that `DeltaV=V_2-0.0709L=1.14L`

`=>V_2=1.14L+0.0709L=1.21L`

Here is a video that further explains this topic:

Thermochemistry | The Nature of Energy.