Question

Calculate the fall of temperature of helium initially 15 degree celsius when suddenly expanded to 8 times its volume given cp by cv is equal to 5 by 3the temp

Answer 1

Here, T1=273+15=288K ,

T2=?,V2=8V1,

γ=5/3

expansion is:

∴T2Vγ−12=T1Vγ−11

∴T2=T1(V1V2)γ−1=288(V18V1)(53−1)

or logT2=log288+23log(18)

log288+23[log1−log8]

logT2=2.4594+23[0−0.90031]

1.8573

T2=antilog1.8573=71.99K

∴ fall in temperature of helium

T1−T2=288−71.99=216.01K

=216.01∘C

(∴in magnitude,1∘C=1K)