JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Heat & Thermodynamics

Following figure shows two processes A and B for a gas. If

Δ Q_{A}

and

Δ Q_{B}

are the amount of heat absorbed by the system in two cases, and

Δ U_{A}

and

Δ U_{B}

are changes in internal energies, respectively, then:

Answer: (B)

Solution

Internal energy is a state function, so since both paths start and end at the same states,

Δ U_{A} = Δ U_{B}

. On the P-V diagram path A encloses the larger area and therefore does more work,

W_{A} > W_{B}

. Applying the first law

Δ Q = Δ U + W

with equal

Δ U

but larger work for A gives

Δ Q_{A} > Δ Q_{B}

while

Δ U_{A} = Δ U_{B}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option