Physics, asked by edwinhmar, 1 month ago

b) For a thermodynamic system, isobaric coefficient of volume expansion (a) and
isothermal compressibility (B) are defined as
1 av
a=
)
и от
p
1 av
B=-
У др
T T
Show that for an isochoric change, ß dp = a dT.​

Answers

Answered by suvendu1274
0

Explanation:

Given,

\alpha = \frac{1}{V}(\frac{dV}{dT})_pα=

V

1

(

dT

dV

)

p

\beta = - \frac{1}{V}(\frac{dV}{dP})_Tβ=−

V

1

(

dP

dV

)

T

Now, from the ideal gas equation,

PV=nRTPV=nRT

V=\frac{nRT}{P}V=

P

nRT

Now, taking the differentiation with respect to the corresponding terms,

\frac{dV}{dT}=\frac{nR}{P}

dT

dV

=

P

nR

if P is constant then,

(\frac{dV}{dT})_P=\frac{nR}{P}...(i)(

dT

dV

)

P

=

P

nR

...(i)

now, taking the differentiation with respect to P,

(\frac{dV}{dP})_T=\frac{nRT}{P^2}...(ii)(

dP

dV

)

T

=

P

2

nRT

...(ii)

Now, taking the ratio of (i)(i) and (ii)(ii)

\frac{(\frac{dV}{dT})_P}{(\frac{dV}{dP})_T}=\frac{\frac{nR}{P}}{\frac{nRT}{P^2}}

(

dP

dV

)

T

(

dT

dV

)

P

=

P

2

nRT

P

nR

\frac{(dP)_T}{(dT)_P}=\frac{P}{T}

(dT)

P

(dP)

T

=

T

P

(dP)_T=(dT)_P\times \frac{P}{T}(dP)

T

=(dT)

P

×

T

P

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