VITEEE 2009 Paper

From first law of thermodynamics
Q = ΔU + W
For cylinder A pressure remains constant.
∴ Work done by a system
W=

µR

γ−1

(T1−T2)
For monoatomic gases, µ=1;γ=

5

3

∴ W=

1×R

5 3 −1

(442−400)=

3

2

R×42
or W = 63 R
But ΔU = 0, for cylinder A
∴ Q = 0 + 63R = 63R
For cylinder B volume is constant,
∴ W=0‌and‌Q=µCvΔT
For monoatomic gas
Cv=

3

2

R ⇒ Q=1×

3

2

RΔT
As heat given on both cylinder is same
∴ 63R=

3

2

RΔT ⇒ ΔT = 42 K