Question

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

A. Calculate the work done by the gas

B. Calculate the change in internal energy of the gas.

Answer 1

A: The work done by a gas undergoing a change in volume is given by
W = PΔV
Since the gas is maintained at atmospheric pressure (1 atm = 1.013 * 10^5 N/m^2),

P = 1.013 * 10^5 N/m^2 and
W = (1.013 * 10^5 N/m^2)(16.2 m^3 - 12.0 m^3)
W = 4.25 * 10^5 J

B: Change in internal energy is given by
ΔU = Q - W (First Law of Thermodynamics), where Q is the heat ADDED to the system, and W is the work done BY the system.

We know that 254 kcal of heat is added to the system and that the system does 4.25* 10^5 J of work,

but we need to convert 1280 kcal to joules before we plug the values in:

254 kcal * 1000 cal/kcal * 4.184 J/cal = 1.06 * 10^6 J

So
ΔU = Q - W
ΔU = 1.06 * 10^6 J - 4.25* 10^5 J
ΔU = 6.3 * 10^5 J