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

Answer:

$V_2=1.363x10^{-3}m^3=1363mL$

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:

$V_2=V_1+\frac{W}{P}$

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

$P=783Torr*\frac{101325Pa}{760Torr} =104394Pa$

And the volumes in m³:

$V_1=67.1mL*\frac{1m^3}{1x10^6mL} =6.71x10^{-5}m^3$

Thus, the final volume turns out:

$V_2=6.71x10^{-5}m^3+\frac{135.3Pa*m^3}{104394Pa}\\\\V_2=1.363x10^{-3}m^3=1363mL$

Best regards.