In the P - V diagram, I is the initial state and F is the final state. The gas goes from I ot F by (i) IAF, (ii) IBF, (iii) ICF. The heat absorbed by the gas is

Question

In the P - V diagram, I is the initial state and F is the final state. The gas goes from I ot F by (i) IAF, (ii) IBF, (iii) ICF. The heat absorbed by the gas is (A) the same in all three processes (B) the same in (i) and (ii) (C) greater in (i) than in (ii) (D) the same in (i) and (iii)

Tardigrade Admin · Accepted Answer

Heat absorbed in a thermodynamic process is given by

Δ Q = Δ U + Δ W

.
Here

Δ U

is same for all the three processes as it depends only on initial and final states.
But

Δ W_{I} = + V e, Δ W_{II} = 0, Δ W_{III} = -

ve

∴ Δ Q_{I} > Δ Q_{II}

Q. In the P−V diagram, I is the initial state and F is the final state. The gas goes from I ot F by (i) IAF, (ii) IBF, (iii) ICF. The heat absorbed by the gas is

the same in all three processes

the same in (i) and (ii)

greater in (i) than in (ii)

the same in (i) and (iii)

Heat absorbed in a thermodynamic process is given by ΔQ=ΔU+ΔW. Here ΔU is same for all the three processes as it depends only on initial and final states. But ΔWI​=+Ve,ΔWII​=0,ΔWIII ​=− ve ∴ΔQI​>ΔQII​

Q. In the $P - V$ diagram, $I$ is the initial state and $F$ is the final state. The gas goes from $I$ ot $F$ by (i) IAF, (ii) IBF, (iii) ICF. The heat absorbed by the gas is

Heat absorbed in a thermodynamic process is given by $Δ Q = Δ U + Δ W$ .
Here $Δ U$ is same for all the three processes as it depends only on initial and final states.
But $Δ W_{I} = + V e, Δ W_{II} = 0, Δ W_{III} = -$ ve
$∴ Δ Q_{I} > Δ Q_{II}$