(A)→(Q) When the value is opened, it is a case of free expansion under a diabetic condition: ΔA = 0, ΔW = 0, ΔU = 0 where T = constant. (B)→(P), (R) We have P =
k
V2
However, PV = nRT ⇒
k
V
= nRT If V → 2V, then T →
T
2
. Therefore. ΔV = n(
3
2
R)ΔT = n(
3
2
R)(−
k
nRV2
)ΔV ΔU =
−3k
2V
Now, ΔW =
2V
∫
V
PdV = k
2V
∫
V
dV
V2
= −k(
1
2V
−
1
V
) =
k
2V
Therefore, ΔQ = ΔU + ΔW =
3k
2V
+
k
2V
= (−
k
V
) < 0 (C)→ (P), (S) P =
k
V
4
3
However, PV = nRT Therefore,
k
V
1
3
= nRT If V → 2V, then T = [
T
(2)
1
3
] < T. ΔV = n(
2
3
R)T(
1
2
1
3
−1) =
3k
2V
1
3
(
1
2
1
3
−1) ΔW =
2V
∫
V
pdV = k
2V
∫
V
dV
V
4
3
= −3k
2V
∫
V
(V)−
1
3
= - 3k (
1
(2V)
1
3
−
1
V
1
3
) Now, ΔQ = ΔU + ΔW =
9k
2V
1
3
(1−
1
2
1
3
) > 0 (D)→(Q), (S) We have PV = nRT That is, P′(2V ) = nRT ′ Therefore,
T′
T
(
2P′
P
) > 1 as P′ > P and T ′ > T Now, ΔW = (Area under PV graph) > 0 ΔU = n(