Question

A heat engine using a diatomic gas follows the cycle shown in the ??pV diagram.

The gas starts out at point 1 with a volume of ?1=233 cm3,V1=233 cm3, a pressure of ?1=147 kPa,p1=147 kPa, and a temperature of 287 K.287 K. The gas is held at a constant volume while it is heated until its temperature reaches 455 K455 K (point 2). The gas is then allowed to expand adiabatically until its pressure is again 147 kPa147 kPa (point 3). The gas is maintained at this pressure while it is cooled back to its original temperature of 287 K287 K (point 1 again).

What is Q₃₁?

Answer 1

As from 3 to 1 process is constant pressure so we can wirte Q31 as

Q31 = nCpΔT

Where Cp for an diatomic gas = 7R/2

Q31 = n(7R/2)(T2-T1) = (7/2)(P1V1 - P3V3)

Pressure at point 2

By process 1 to 2

This is a constant volume process

V = Constant

T/P = Constant

287/147 = 455/P2

P2 = 888.33 kPa

Volume of point 3

By the adiabatic relation

P(V)^γ = Constant

γ for diatomic gas = 1.4

So 888.33(233)1.4 = 147(V)1.4​​​​​​

V = 842.67 Cm^3

So Q31 = (7/2)(P1V1 - P3V3) = (7/2)(147×233 - 147×842.67)×103×10-6 = - 2.13 J