One mole of a monatomic ideal gas undergoes the cyclic process J K L M J, as shown in the P-T diagram. [figure] Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [ R is the gas constant.]
- P → 1; Q → 3; R → 5; S → 4
- P → 4; Q → 3; R → 5; S → 2
- P → 4; Q → 1; R → 2; S → 2
- P → 2; Q → 5; R → 3; S → 4
Answer
(2) P → 4; Q → 3; R → 5; S → 2
Solution
States: J(P0,T0), K(P0,3T0), L(2P0,3T0), M(2P0,T0); n=1, CV= 32 R.
JK (isobaric): W= R(2T0)=2 RT0, U= 32 R(2T0)=3 RT0 → Q → 3.
KL (isothermal at 3T0, P doubled): W=3 RT0 12=-3 RT0 2=QKL → R → 5.
LM (isobaric): W= R(T0-3T0)=-2 RT0. MJ (isothermal at T0, P halved): W= RT0 2, U=0 → S → 2.
Net work =2 RT0-3 RT0 2-2 RT0+ RT0 2=-2 RT0 2 → P → 4.
Answer: (B)