BITSAT 2018 Slot 2 Solved Paper

Given, C=37.55Jmol−1K−1
Also, pT=/constant(K) ........(i)
According to ideal gas equation, pV=RT
⇒p=

RT

V

...........(ii)
Putting the value of p in Eq. (i), we get

RT

V

×T=K⇒V=

RT2

K

On differentiating above equation both sides, we get

dV

dT

=

2RT

K

........(iii)
But

T

K

=

1

p

[from Eq. (i)]
Hence, Eq. (iii) becomes

dV

dT

=

2R

p

So, C=CV+

pdV

dT

orC=CV+

p×2R

p

=CV+2R
CV=C−2R .........(iv)
or As CV=

nR

2

, where n= number of degrees of freedom.
Putting the value of CV in Eq. (iv), we get

nR

2

=C−2R
n=

2(C−2R)

R

=

2(37.55−2×8.3)

8.3

=5.048=≃5