Question

A gas is confined to a container with a massless piston at the top. (Figure 2)  A massless wire is attached to the piston. When an external pressure of 2.03 bar is applied to the wire, the gas compresses from 6.40 to 3.20 L . When the external pressure is increased to 2.53 bar, the gas further compresses from 3.20 to 2.56 L .

In a separate experiment with the same initial conditions, a pressure of 2.53 bar was applied to the gas, decreasing its volume from 6.40 to 2.56 L in one step.

If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

Express your answer with the appropriate units.

Answer 1

for two step process :

pressure = 2.03 bar

work done = - p dV

                 = - 2.03 (3.20 - 6.40 )

                 = 6.496 L.atm

q1 = 6.496 L.atm

work = - 2.53 (2.56 - 3.20) = 1.6192 L.atm

q2 = 1.6192 L.atm .

total q = 6.496 + 1.6192 = 8.1152

for one step process :

work = - 2.53 (2.56 - 6.40) = 9.7152 L.atm

q = 9.7152 L.atm

difference = 8.1152 - 9.7152 = - 1.6 L.atm

difference in q = 162 J