Question

A mole of helium at 20°C is suddenly subjected to a pressure of 10 times its initial value. Find the final temperature attained
solution​

Answer 1

Explanation:

The PV diagram for the given changes is as:

at A: T

1

=300K,n=2,V=20litre,P=P

1

at B: T=T

2

,n=2,V=40litre,P=P

1

at C: T=300K.n=2,V=V

1

litre,P=P

2

For A P

1

=

V

nRT

=

20

2×0.082×300

=2.46atm

For isobaric process AB:

Also

T

1

V

1

=

T

2

V

2

(fromCharles

′

lawatconstantP)

300

20

=

T

2

40

∴T

2

=600K

For adiabatic process BC:

TV

y−1

=constant

600×(40)

y−1

=300×(V

1

)

y−1

or (

40

V

1

)

(

3

5

1)

=2

or (

40

V

1

)

2/3

=2

∴V

1

=113.13litre

The work done W

T

=W

AB

+W

BC

W

AB

:∴w=nRT.dT

w=2×2×(600−300)=1200cal

W

BC

:foradiabaticprocess;

∴w=−nC

v

.ΔT

=−2×

2

3

×R×(600−300)

=1800cal

∴W

r

=1200+1800=3000cal