JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Heat & Thermodynamics

For a gas

C_{P} - C_{V} = R

in a state

P

and

C_{P} - C_{V} = 1.10 R

in a state

Q

T_{P}

and

T_{Q}

are the temperatures in two different states P and Q respectively. Then

(A) $T_{P} < T_{Q}$
(B) $T_{P} = 0.9 T_{Q}$
(C) $T_{P} > T_{Q}$ ✓ Correct
(D) $T_{P} = T_{Q}$

Answer: (C)

Solution

For an ideal gas,

C_{P} - C_{V} = R

always holds. When

C_{P} - C_{V} = 1.10 R

, the gas is behaving as a real (non-ideal) gas. Real gases approach ideal behavior at high temperatures and low pressures. Since state P satisfies the ideal gas relation (

C_{P} - C_{V} = R

) while state Q shows deviation (

C_{P} - C_{V} = 1.10 R

), state P must be at a higher temperature where the gas behaves more ideally. Therefore,

T_{P} > T_{Q}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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