Question

A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0m3 to 16.2m3 .

a. Calculate the work done by the gas.

b. Calculate the change in internal energy of the gas.]

Express your answer with the appropriate units.

Answer 1

Work done in an isobaric process:
W = P*(V2 - V1)

First law equation:
Q = ?U + W

Substitute and solve for Q:
?U = Q - P*(V2 - V1)

Data:
P:=101.325 kPa; V2:=16.2 m^3; V1:=12.0 m^3; Q:=7117.56 kJ;

Results:
Work: W = 425.6 kJ
Internal energy change: ?U = 6692 kJ

Comment: the flavor of gas must involve a really complicated molecule, because from k = Q/?U, I calculated that the adiabatic index is very close to one, k=1.064.