A gas is confined to a container with a massless piston at the top. A massless w

Question

A gas is confined to a container with a massless piston at the top. 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 4.40 to 2.20L . When the external pressure is increased to 2.50 atm, the gas further compresses from 2.20 to 1.76L . In a separate experiment with the same initial conditions, a pressure of 2.50 atm was applied to the gas, decreasing its volume from 4.40 to 1.76L 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?

Explanation / Answer

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 = 2.5·101325 Pa · (4.40×10?³ m³ - 1.76×10?3 m³) = 668.745 Pa·m3 = 668.7 J For the two-step process W2 = 2.0·101325 Pa · (4.40×10?³ m³ - 2.20×10?3 m³) + 2.5·101325 Pa · (2.20×10?³ m³ - 1.76×10?3 m³) = 557.3 J Therefore ?Q = W1 - W2 = 668.7 J - 557.3 J = 111.4

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