Question

An ideal gas (which is is a hypothetical gas that conforms to the laws governing gas behavior) 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 2.00 atm is applied to the wire, the gas compresses from 5.40 to 2.70 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.70 to 2.16 L . In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the ideal gas, decreasing its volume from 5.40 to 2.16 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?

Answer 1

According to First law of thermodynamics,

deltaU = q + W

But, temperature is remained constant, deltaU = 0

So, q = - W

But, pressure - volume work can be defiend as,

W = - P * delta V

Therefore,

q = P * deltaV

Case(1)

(a) two step process,

q1 = 2.00 * (5.40 - 2.70) = 5.40 L.atm

q2 = 2.50 * (2.70 - 2.16) = 1.35 Latm.

Total heat change, q = 5.40 + 1.35 = 6.75 L.atm

(b) one step process,

q = 2.50 * (5.40 - 2.16) = 8.10 L.atm

Therefore, heat difference in both process = 8.10 - 6.75 = 1.35 L.atm = 1.35 * 101.325 J = 137. J