Question

A sample of gas occupies a volume of 67.1 mL . As it expands, it does 135.3 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=1.363x10^(-3)m^3=1363mL`

Step-by-step explanation:

Hello,

In this case, since the work done at constant pressure as in isobaric process is computed by:

`W= P(V_2-V_1)`

Thus, given the pressure, initial volume and work, the final volume is:

`V_2=V_1+(W)/(P)`

Whereas the pressure must be expressed in Pa as the work is given in J (Pa*m³):

`P=783Torr*(101325Pa)/(760Torr) =104394Pa`

And the volumes in m³:

`V_1=67.1mL*(1m^3)/(1x10^6mL) =6.71x10^(-5)m^3`

Thus, the final volume turns out:

`V_2=6.71x10^(-5)m^3+(135.3Pa*m^3)/(104394Pa)\\\\V_2=1.363x10^(-3)m^3=1363mL`

Best regards.