Problem 2

One mole of monatomic ideal gas starts at state

A

with

T_A=300

K and

V_A=0.010

^3

. It follows a three-step cycle: (1) heat at constant volume from

A

B

, where

T_B=600

K; (2) expand isothermally from

B

C

until the volume doubles; (3) compress at constant pressure from

C

back to

A

. Use

R=8.31

J mol

^{-1}

^{-1}

, molar

C_V=3R/2

, and define work

W

as positive when done by the gas. (a) Find the pressures at

A,B,C

. (b) Calculate work and heat for each leg. (c) Check the first law over the complete cycle.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

W_{CA}=P_A(V_A-V_C)=-P_AV_A=-nRT_A=-2493\ \mathrm J,\quad \Delta U_{CA}=\frac32R(300-600)=-3739.5\ \mathrm J,\quad Q_{CA}=-6232.5\ \mathrm J

problems.proof.analysis

Pressure is constant while volume falls back to $V_A$ , so work done by the gas is negative. The temperature drop lowers internal energy; $Q=\Delta U+W$ gives the heat rejected.