Question

Show that coefficient of cubical expansion of an ideal gas at constant pressure is equal to
its absolute temperature

Answer 1

Given:

Cubical expansion Coefficient of ideal gas at the constant pressure is equal to absolute temperature

To Find:

Prove the above statement

Solution:

We know from cubical expansion formula

Cubical expansion coefficient (Υ) = $\dfrac{1}{V} \dfrac{(dv)}{(dt)_{p} }$

From an ideal gas

PV = n RT

V = Volume

P = Pressure

n = moles

R = gas constant

T = Temperature

$T = \dfrac{PV}{nR}$

By differentiating this equation-

$PdV + VdP = nRdT$

At constant pressure dP = 0

Then

$PdV = nRdT$

$\dfrac{dV}{dT} = \dfrac{nR}{P}$

Equate this equation with cubical coefficient expansion

Υ = $\dfrac{1}{V} \dfrac{nR}{P}$

$T = \dfrac{PV}{nR}$

Υ = $\dfrac{1}{T}$

Hence proved

Answer 2

Explanation:

We know from cubical expansion formula

Cubical expansion coefficient (Υ) = \dfrac{1}{V} \dfrac{(dv)}{(dt)_{p} }

V

1

(dt)

p

(dv)

From an ideal gas

PV = n RT

V = Volume

P = Pressure

n = moles

R = gas constant

T = Temperature

T = \dfrac{PV}{nR}T=

nR

PV

By differentiating this equation-

PdV + VdP = nRdTPdV+VdP=nRdT

At constant pressure dP = 0

Then

PdV = nRdTPdV=nRdT

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

dT

dV

=

P

nR

Equate this equation with cubical coefficient expansion

Υ = \dfrac{1}{V} \dfrac{nR}{P}

V

1

P

nR

T = \dfrac{PV}{nR}T=

nR

PV

Υ = \dfrac{1}{T}

T

1

Hence proved