A gas is expanded from volume V0 to 2V0 under three different processes, as shown in the figure. Process 1 is 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 expansion (P = constant) Hence temperature of gas will increase. ∴ Δ U1 = positive Process 2 is an isothermal process ∴ 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, as shown in the figure. Process 1 is 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 expansion (P= constant) Hence temperature of gas will increase. ∴ΔU1​= positive Process 2 is an isothermal process ∴ΔU2​=0 Process 3 is an adiabatic expansion Hence temperature of gas will fall. ∴ΔU3​= negative ∴ΔU1​>ΔU2​>ΔU3​

$Δ U_{1} > Δ U_{2} > Δ U_{3}$

$Δ U_{1} < Δ U_{2} < Δ U_{3}$

$Δ U_{2} > Δ U_{1} > Δ U_{3}$

$Δ U_{2} < Δ U_{3} < Δ U_{1}$