A gas is expanded from volume V0 to 2V0 under three different processes. In figure process 1 is an isobaric process, process 2 is isothermal and process 3 is adiabatic. Let Δ U1, Δ U2 and Δ U3 be the change in internal energy of the gas in these three processes. Then

Question

A gas is expanded from volume V0 to 2V0 under three different processes. In figure process 1 is an isobaric process, process 2 is isothermal and process 3 is adiabatic. Let Δ U1, Δ U2 and Δ U3 be the change in internal energy of the gas in these three processes. Then Question

(A) Δ U1>Δ U2>Δ U3 (B) Δ U1 < Δ U2 < Δ U3 (C) Δ U2 < Δ U1 < Δ U3 (D) Δ U2 < Δ U3 < Δ U1

Tardigrade Admin · Accepted Answer

Process 1 is isobaric (.p= constant) expansion Hence, temperature of gas will increase ∴ Δ U1= positive Process 2 is an isothermal expansion ∴ Δ U2=0 Process 3 is an adiabatic expansion Hence, temperature of gas will fall ∴ Δ U3= negative ∴ Δ U1>Δ U2>Δ U3

$Δ U_{1} > Δ U_{2} > Δ U_{3}$

$Δ U_{1} < Δ U_{2} < Δ U_{3}$

$Δ U_{2} < Δ U_{1} < Δ U_{3}$

$Δ U_{2} < Δ U_{3} < Δ U_{1}$

Q. A gas is expanded from volume V0​ to 2V0​ under three different processes. In figure process 1 is an isobaric process, process 2 is isothermal and process 3 is adiabatic. Let ΔU1​,ΔU2​ and ΔU3​ be the change in internal energy of the gas in these three processes. Then

ΔU1​>ΔU2​>ΔU3​

ΔU1​<ΔU2​<ΔU3​

ΔU2​<ΔU1​<ΔU3​

ΔU2​<ΔU3​<ΔU1​

Process 1 is isobaric (p= constant) expansion Hence, temperature of gas will increase ∴ΔU1​= positive Process 2 is an isothermal expansion ∴ΔU2​=0 Process 3 is an adiabatic expansion Hence, temperature of gas will fall ∴ΔU3​= negative ∴ΔU1​>ΔU2​>ΔU3​

$Δ U_{1} > Δ U_{2} > Δ U_{3}$

$Δ U_{1} < Δ U_{2} < Δ U_{3}$

$Δ U_{2} < Δ U_{1} < Δ U_{3}$

$Δ U_{2} < Δ U_{3} < Δ U_{1}$