Thermodynamics Part 3

Given,
Change in pressure with volume

dp

dV

=−ap
At V=0,p=p0
Number of mole n = 1
Gas constant, R=8.314JK−1
From given equation,

dp

p

=−adV

p

∫

p0

dp

p

=−a‌

v

∫

0

dV
⇒[ln‌p]p0p=−a[V]0V
⇒ln(

p

p0

)=−a(V−0)
Taking anti log on both side,
⇒

p

p0

=e−aV
⇒p=p0e−aV
By using ideal gas law,
pV = nRT
⇒(p0e−aV)V=nRT
⇒

1

nR

(p0e−aV)V=T
⇒

1

R

(p0e−aV)V=T [∵ n = 1]…(i)
For maximum temperature (Tmax),
On differentiating both side with respect to V, we get
⇒

1

nR

[p0.e−aV.(−aV)V+p0e−aV]=

dT

dV

=0
⇒p0e−aV(1−aV2)=0
⇒1=aV2
⇒V2=

1

a

⇒V=

1

a

Substituting the value in Eq. (i), we get
T=

1

R

(p0e−a

.1

a

)

1

a

=

p0e−1

R

(

1

a

)=

p0

eRa