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 6.40 to 3.20 L . When the external pressure is increased to 2.50 atm, the gas further compresses from 3.20 to 2.56 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 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?

Answer 1

Note that internal energy is a state function. That means internal energy of the gas can be expressed as function of two state variables, e.g. U = f(T;V). For an ideal gas internal function can be expressed of temperature alone. But is not necessary to make ideal gas assumption to solve this problem.
Because internal energy is a state function, a process changing from state 1 to state 2 has always the same change change in internal energy irrespective of the process design.
The one-step compression and the two two-step compression start at the sam state and end up in same state. the gas undergoes the same change in internal energy:
∆U₁ = ∆U₂
The change in internal energy of the gas equals the heat added to the gas plus work done on it:
Hence,
Q₁ + W₁ = Q₂ + W₂
So the difference in heat transfer between the two process is:
∆Q = Q₂ - Q₁ = W₁ - W₂

The work done on the gas is given by piston is given by the integral
W = - ∫ P_ex dV from V_initial to V_final
For constant external pressure like in this problem this simplifies to
W = - P_ex ∙ ∫ dV from V_initial to V_final
= P_ex ∙ (V_initial - V_final)

The work done in one step process is:
W₁ = 2.5*1.01*10⁵ Pa * (6.40×10^-3m³ - 2.56×10^-3 m³)
= 969.6Pa∙m3
= 969.6 J
For the two-step process
W₂ = 2.0*1.01*10⁵ Pa *(6.40×10^-3 m³ - 3.20×10^-3 m³)
+ 2.5*1.01*10⁵ Pa *(3.20×10^-3 m³ - 2.56×10^-3 m³)
= 808 J

Therefore
∆Q = W₁ - W₂
= 969.6 J - 808 J
= 161.6J