Answer:
[tex]V_2=1.363x10^{-3}m^3=1363mL[/tex]
Explanation:
Hello,
In this case, since the work done at constant pressure as in isobaric process is computed by:
[tex]W= P(V_2-V_1)[/tex]
Thus, given the pressure, initial volume and work, the final volume is:
[tex]V_2=V_1+\frac{W}{P}[/tex]
Whereas the pressure must be expressed in Pa as the work is given in J (Pa*m³):
[tex]P=783Torr*\frac{101325Pa}{760Torr} =104394Pa[/tex]
And the volumes in m³:
[tex]V_1=67.1mL*\frac{1m^3}{1x10^6mL} =6.71x10^{-5}m^3[/tex]
Thus, the final volume turns out:
[tex]V_2=6.71x10^{-5}m^3+\frac{135.3Pa*m^3}{104394Pa}\\\\V_2=1.363x10^{-3}m^3=1363mL[/tex]
Best regards.
