Column-I contains different processes undergone by a diatomic ideal gas. Column-II change in different parameter of ideal gas.
Column-I | Column-I |
(i) PV–1 = constant and volume is increased twice | [A] Heat is given to gas |
(ii) P2V = constant and pressure is increased twice | [B] Heat is rejected by gas |
(iii) PV6/5 = constant and volume is reduced to half the initial volume | [C] Work done by gas is negative |
(iv) PV2 = constant and pressure is increased 3 times | [D] Internal energy increase |
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Ans.
(i) [A],[D]
(ii) [B],[C]
(iii) [B],[C],[D]
(iv) [A],[B],[C]
Sol.
For process PV n = constant
Molar heat capacity of gas
C = R 
... (i)
Here, γ = 7/5
PV = nRT = 
= Constant × 
For n = 1 : Temperature with increase in
volume work done positive
Hence heat is absorbed by system.
For n = 
: Temperature and volume decrease
with increase in pressure
∴ Work done negative
Hence heat is rejected
For n = 
: Temperature increases with decrease
in volume work done negative
For n = 2 : Temperature increase with increase in pressure work done negativeHence heat is absorbed.
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