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

Following figure shows two processes

A

and

B

for a gas. If

∆ {Q}_{A}

and

∆ {Q}_{B}

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

∆ {U}_{A}

and

∆ {U}_{B}

are changes in internal energies, respectively, then:
Following figure shows two processes A and B for a gas. If ∆ Q _ A and ∆ Q _ B are the amount of heat absorbed by the

Following figure shows two processes A and B for a gas. If ∆ Q _ A and ∆ Q _ B are the amount of heat absorbed by the

Options

A $∆ {Q}_{A} < ∆ {Q}_{B}; ∆ {U}_{A} < ∆ {U}_{B}$
B $∆ {Q}_{A} = ∆ {Q}_{B}; ∆ {U}_{A} = ∆ {U}_{B}$
C $∆ {Q}_{A} > ∆ {Q}_{B}; ∆ {U}_{A} = ∆ {U}_{B}$
D $∆ {Q}_{A} > ∆ {Q}_{B}; ∆ {U}_{A} > ∆ {U}_{B}$

Correct answer

C. $∆ {Q}_{A} > ∆ {Q}_{B}; ∆ {U}_{A} = ∆ {U}_{B}$

Step-by-step solution

⇒ ∆ {U}_{A} = ∆ {U}_{B}

as change in internal energy depends on initial and final state and not the path.

⇒

By comparing area under the curve,

{W}_{{}_{A}} > {W}_{{}_{B}}

From

Q = U + W

We can say,

{Q}_{{P}_{A}} > {Q}_{{P}_{B}}

Practice Thermodynamics on Quantrex Academy →

More from Thermodynamics