An ideal monatomic gas undergoes different types of processes which are described in column-I match the corresponding effects in column-II. The letters have usual meaning.
Column-I | Column-II |
(i) P=2V2 | [A] If volume increases then temperature will also increase |
(ii) PV2=constant | [B] If volume increases then temperature will decrease |
(iii) C=CV +2R | [C] For expansion, heat will have to be supplied to the gas |
(iv) C=CV–2R | [D] If temperature increases then work done by gas in positive |
Text Solution
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Ans.
(i) [A], [C], [D]
(ii) [B]
(iii) [A], [C], [D]
(iv) [B], [C]
Sol.
(i) If P = 2V 2 , from ideal gas equation we get 2V 3 = nRT
∴ with increase in volume
(i) Temperature increases implies dU = + ve
(ii) dw = + veHence dQ = dU + dw = + ve
(ii) If PV 2 = constant, from ideal gas equation we get VT = k (constant)
Hence with increase in volume, temperature decreases
Now dQ = dU + PdV = nCvdT –PK/T 2 dT [ 
dV = – K/T 2 dT]
= nC V dT –PV/T dT = n(C v –R) dT
∴ with increase in temperature dT = +ve and since C V >R for monoatomic gas. Hence dQ = +ve as temperature is increased
(iii) dQ = nC dT = nC v dT + PdV
⇒ n(C v + 2R) dT = nC v dT + PdV
∴ 2nRdT = PdV ∴ dV/dT = +ve
Hence with increase in temperature volume increase and vice versa
∴ dQ = dU + dw = +ve
(iv) dQ = nC dT = nC v dT = PdV
or n (C v –2R)dT = nC v dT + PdV
or –2nRdT = PdV ∴ dV/dT = –ve
∴ with increase in volume temperature decreases. Also dQ = n(C v –2R) dT
with increase in temperature dT = + ve but C v < 2R for monoatomic gas. Therefore
dQ = –ve with increase in temperature.
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