Question

in the first part of two step process heat is allowed to flow out of an ideal gas at constant volume so that the presure drop from 2.45atm to 1.70atm the gast the expands at constant pressure from a volume of 5.45l to 7.85L where the tempetrture rekbei it original value .

a draw a pv -disgram for this process

b , calculate the total work done bu the gas process

c does heat flow into or out of the gas dutring this process ,what is the amount of heat

Answer 1

(a)
PV Diagram -



(b)
The total work done is the sum of the work done in each step.
For Process 1 -
W = - ∫ p dV from initial to final volume

Since the the volume does not change in first step, no work is done on the gas:
W₁ = 0

In second step gas expands at constant pressure, so the work integral simplifies to:
W₂ = p ∙ ∫ dV from initial to final volume
W₂ = p ∙ (V_final - V_initial)
W₂ = 1.7 * 101325 Pa ∙ (0.00785 m³ - 0.00545 m³)
W₂ = 413.4 J

So the total work in the two-step process is:
W = W₁ + W₂ = 0J + 413.4
W = 413.4 J


(c)
Assuming ideal gas behavior, change of internal energy is guven by:
∆U = n∙Cv∙∆T

Because we return to initial temperature after second step, the change of internal energy is zero
∆U = 0

Change of internal energy equals work done On the gas plus heat transferred to the gas:
∆U = W + Q
Q = ∆U - W = 0J - (413.4 J) = - 413.4 J

This means 413.4 J of heat is flowing into the gas during the two-step process.