GSO001 Problem 7
Problem Statement
Two-Stage Process and Heat Cycle for a Monatomic Ideal Gas
Problem Statement
A cylinder with a frictionless piston contains n moles of a monatomic ideal gas. Initially, the gas has pressure P0, volume V0, and temperature T0 (state A). The following two processes are performed in sequence:
- Process 1 (Isochoric heating): The gas is heated at constant volume V0 until the pressure reaches 2P0 (state B).
- Process 2 (Isobaric expansion): The gas is heated at constant pressure 2P0 until the volume reaches 2V0 (state C).
Let R be the gas constant and the molar heat capacity at constant volume of the monatomic ideal gas be CV=23R. Find the total heat Q absorbed by the gas from outside during the combined process (processes 1 and 2).
Constraints
- Initial pressure: P0=2.0×103 Pa
- Initial volume: V0=0.50 m3
- Gas constant: R=8.3 J/(mol\cdotK) (you may use the relation P0V0=nRT0 in calculations)
Input Format
Find the value of Q and give the integer part of the answer.
Solution
This is a standard problem applying the first law of thermodynamics Q=ΔU+W to each process and summing the total heat.
1. Process 1 (A → B: Isochoric Change)
In an isochoric process, the work done by the gas is W1=0. So the heat absorbed Q1 equals the change in internal energy ΔU1.
For a monatomic ideal gas, the change in internal energy is ΔU=23nRΔT=23Δ(PV).
Q1=ΔU1=23(2P0V0−P0V0)=23P0V02. Process 2 (B → C: Isobaric Change)
In an isobaric process, the work done by the gas W2=PΔV:
W2=2P0(2V0−V0)=2P0V0The change in internal energy ΔU2:
ΔU2=23(2P0⋅2V0−2P0⋅V0)=23(2P0V0)=3P0V0Heat absorbed in process 2:
Q2=ΔU2+W2=3P0V0+2P0V0=5P0V03. Total Heat
Q=Q1+Q2=23P0V0+5P0V0=6.5P0V0Substituting numerical values:
Q=6.5×(2.0×103)×0.50=6.5×1000=6500 JAnswer: 6500