JEE Main 2019 — Physics solution

JEE Main2019Physics

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:

A
$Δ Q_{A} > Δ Q_{B}, Δ U_{A} = Δ U_{B}$
B
$Δ Q_{A} = Δ Q_{B}, Δ U_{A} = Δ U_{B}$
C
$Δ Q_{A} > Δ Q_{B}, Δ U_{A} > Δ U_{B}$
D
$Δ Q_{A} < Δ Q_{B}, Δ U_{A} < Δ U_{B}$

Solution

$Δ W_{A} > Δ W_{B} Δ U_{A} > Δ U_{B}$

From first law of thermodynamics,

$Δ Q = Δ U + Δ W Δ Q_{A} > Δ Q_{B}$

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