Following figure shows two processes A and B for a gas. If Δ QA and Δ QB are the amount of heat absorbed by the system in two cases, and Δ UA and Δ UB are changes in internal energies, respectively, then :

Question

Following figure shows two processes A and B for a gas. If Δ QA and Δ QB are the amount of heat absorbed by the system in two cases, and Δ UA and Δ UB are changes in internal energies, respectively, then : (A) Δ QA = Δ QB ; Δ UA = Δ UB (B) Δ QA > Δ QB ; Δ UA = Δ UB (C) Δ QA = Δ QB ; Δ UA > Δ UB (D) Δ QA < Δ QB ; Δ UA < Δ UB

Tardigrade Admin · Accepted Answer

Initial and final states for both the processes are same. ∴ Δ UA = Δ UB Work done during process A is greater than in process B. By First Law of thermodynamics Δ Q = Δ U + W ⇒ Δ QA > Δ QB

Q. 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 :

$Δ Q_{A} = Δ Q_{B}; Δ U_{A} = Δ U_{B}$

$Δ Q_{A} > Δ Q_{B}; Δ U_{A} = Δ U_{B}$

$Δ Q_{A} = Δ Q_{B}; Δ U_{A} > Δ U_{B}$

$Δ Q_{A} < Δ Q_{B}; Δ U_{A} < Δ U_{B}$

Initial and final states for both the processes are
same.
$∴ Δ U_{A} = Δ U_{B}$
Work done during process $A$ is greater than in
process $B$ .
By First Law of thermodynamics
$Δ Q = Δ U + W$
$\Rightarrow Δ Q_{A} > Δ Q_{B}$

Q. Following figure shows two processes A and B for a gas. If ΔQA​ and ΔQB​ are the amount of heat absorbed by the system in two cases, and ΔUA​ and ΔUB​ are changes in internal energies, respectively, then :

ΔQA​=ΔQB​;ΔUA​=ΔUB​

ΔQA​>ΔQB​;ΔUA​=ΔUB​

ΔQA​=ΔQB​;ΔUA​>ΔUB​

ΔQA​<ΔQB​;ΔUA​<ΔUB​