Question

A cylinder with a piston contains 0.140 mol of nitrogen at 1.88 105 Pa and 325 K. The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. PLEASE do B and C and don't don't forget to show all the steps specially for B becuase I don't know how to do it. I have the anwers below.

(a) Compute the work done by the gas, the heat added to it, and its internal-energy change during the initial compression.

W = -189 J
Q =   -661 J
ΔU = -472 J

(b) Compute the work done by the gas, the heat added to it, and its internal-energy change during the adiabatic expansion.

W = 115 J
Q = 0 J
ΔU =-115 J

(c) Compute the work done, the heat added, and the internal-energy change during the final heating.

W = 0 J
Q = 587 J
ΔU = 587 J

Answer 1

From the given question,

From first law of thermodynamics

Q= ΔU + W

where q= heat exchange

ΔU = change in internal energy

W=work done

n=0.140

p1=1.88 x 10⁵ Pa and

T1=325 K

R= gas constant=8.314

pv =nRT

v1= (0.14)(8.314)(325)/(1.88 x 10⁵)=0.002m³

1 -> 2 isobaric process( constant pressure)

p2= 1.88 x 10⁵ Pa

v2=v1/2=0.002/2m³= 0.001m³

w= p $\Delta$ v= 1.88 x 10⁵ x 0.001=189 J

v1/T1=v2/T2

v2/v1=T2/T1

1/2=T2/325

T2=162.5K

$\gamma =1.4$

Cv= R/( $\gamma -1$ )=20.785

ΔU= nCv dT=0.14 (20.785)(162.5)=472.85

Q= ΔU + W

=472.85 + 189=661.8

B . During the adiabatic process, there is no heat exchange.

so Q = 0 J

Q= ΔU + W

ΔU = -W

v3=v1=0.002

W=P₂V₂ ^$\gamma$ (V₃^{(1- $\gamma$ )}-V₂^{(1- $\gamma$ )})/(1- $\gamma$ )

=1.88 x 10⁵ x0.001^1.4(0.002^0.4-0.001^0.4)/(1-1.4)

=115

ΔU = -W=-115

In constant volume process

W=0

Q= ΔU + W

Q= ΔU

Q= Cv V (P₃-P₁)/R

(p1/p2)=(v2/v1) ^$\gamma$

p2=71238 pa

Q=20.785 (0.001)(1.88 x 10⁵-71238)/8.314

=587J

Q= ΔU

ΔU=587J