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Question

An ideal gas (molar specific heat CV = 5R2) is taken along paths acb, adb and ab. P2 = 2P1, V2 = 2V1. Along ab, P = kV where k is a constant. The various parameters are shown in figure. Match the column I with the corresponding option of column II.

Column I Column II
i. Wacb p 15RT12
ii. Wadb q 15RT12
iii. Uab r RT1
iv. Ubca s 2RT1

Now match the given columns and select the correct option from the codes given below.

A
i - r, ii - s, iii - p, iv - q
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B
i - s, ii - r, iii - p, iv - q
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C
i - r, ii - s, iii - q, iv - p
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D
i - s, ii - r, iii - q, iv - p
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Solution

The correct option is B i - s, ii - r, iii - p, iv - q
Wacb = Wac + Wcb = 0 + P2(V2 V1) = P2V1 = 2P1V1 = 2RT1( V2 = 2V1)Wadb = Wad + Wdb = P1(V2 V1) + 0= P1V1 = RT1Uab = Uac + Ucb = (Qac Wac) + (Qcb Wcb) = CV(Tc T1) + CV(T2 TC)= CV (T2 T1)= 5R2 (T2 T1)(Given CV =5R2)For an ideal gas,P1V1T1 = P2V2T2P2 = 2P1 and V2 = 2V1 T2 = 4T1Uab = 5R2 (T2 T1) = 15RT12 For the entire cycle, ΔU = 0Ubca = Uab Ubca =15RT12

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