Question

He gas is confined to a piston and cylinder with a mechanical stop that prevents the cylinder from expanding. The temperature of the gas is 298 K, the volume of gas in the cylinder is 1.20 L and its pressure is 2.15 bar. The ideal gas EOS is very accurate for He under these conditions.

Starting from the final state of the He gas in part (b), the gas (and apparatus) was cooled from 298 K to 263 K. The gas in the cylinder was free to change volume, but the external pressure was held at 1.12 bar. The heat capacity of He gas at constant pressure is 20.786 J K-1 mol-1.

a) What is w (in J)? Is the system (He gas in cylinder) doing work on the surroundings, or is the surroundings doing work on the system?
b) What is q (in J)? Is the system gaining or losing energy to the surroundings through the heat mechanism?
c) What is the change in internal energy for the system? What is the change in the energy of the surroundings?

Answer 1

PV=nRT

n =( 2.15/1.01325) atm× 1.20L/(0.0821atm-L/K.mol × 298K )

=0.10407mol

Final volume of system = nRT/P = 0.10407 × 0.0821×263/(1.12/1.01325) = 2.033L

(a)

Work done = - P_ext(V₂ -V₁) = - 1.12bar(2.033L - 1.20L)

= - 0.93298 bar-L = -0.93298×100J = 93.298J

= -93.3 J (answer)

Since final volume of He gas is more than initial volume, means the system has done work on the surroundings .

(b)

Heat capacity at constant pressure = 20.786J/K.mol

Heat lost = n.Cp.∆T = 0.10407 ×20.786×(263-298)

= -75.72 J

Since final temperature is less than that of initial temperature, heat has lost to the surrounding.

(c)

∆U = q +w = - 75.72 J - 93.3J = - 169.02 J

The change in internal energy of system = -169J (answer)

The change in energy of surroundings = 169 J