An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. If Q1,Q2,Q3 indicate the heat absorbed by the gas along the three processes and Δ U1, Δ U2, Δ U3 indicate the change in internal energy along the three processes respectively, then

Question

An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8af58691133be165cbac1f6e95e1ce9f-.jpeg" /> If Q1,Q2,Q3 indicate the heat absorbed by the gas along the three processes and Δ U1, Δ U2, Δ U3 indicate the change in internal energy along the three processes respectively, then (A) Q1>Q2>Q3 and Δ U1=Δ U2=Δ U3  (B) Q3>Q2>Q1 and Δ U1=Δ U2=Δ U3  (C) Q1=Q2=Q3 and Δ U1>Δ U2>Δ U3  (D) Q3>Q2>Q1 and Δ U1>Δ U2>Δ U3

Tardigrade Admin · Accepted Answer

Change in internal energy is path independent and depends only on the initial and final states.
As the initial and final states in the three processes are same. Therefore

Δ U_{1} = Δ U_{2} = Δ U_{3}

Workdone,

W =

Area under

P - V

graph
As area under curve

1 >

area under curve

2 >

area under curve

3

∴ W_{1} > W_{2} > W_{3}

According to first law of thermodynamics,

Q = W + Δ U

As

W_{1} > W_{2} > W_{3}

and

Δ U_{1} = Δ U_{2} = Δ U_{3}

∴ Q_{1} > Q_{2} > Q_{3}

$Q_{1} > Q_{2} > Q_{3}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$

$Q_{3} > Q_{2} > Q_{1}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$

$Q_{1} = Q_{2} = Q_{3}$ and $Δ U_{1} > Δ U_{2} > Δ U_{3}$

$Q_{3} > Q_{2} > Q_{1}$ and $Δ U_{1} > Δ U_{2} > Δ U_{3}$

Q1​>Q2​>Q3​ and ΔU1​=ΔU2​=ΔU3​

Q3​>Q2​>Q1​ and ΔU1​=ΔU2​=ΔU3​

Q1​=Q2​=Q3​ and ΔU1​>ΔU2​>ΔU3​

Q3​>Q2​>Q1​ and ΔU1​>ΔU2​>ΔU3​

$Q_{1} > Q_{2} > Q_{3}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$

$Q_{3} > Q_{2} > Q_{1}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$

$Q_{1} = Q_{2} = Q_{3}$ and $Δ U_{1} > Δ U_{2} > Δ U_{3}$

$Q_{3} > Q_{2} > Q_{1}$ and $Δ U_{1} > Δ U_{2} > Δ U_{3}$