Question 11

NUMERICALHARD

An ideal monatomic gas of nn moles is taken through a cycle WXYZWWXYZW consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V−TV-T diagram. The volume of the gas at W,XW, X and YY points are, 64 cm364\text{ cm}^3, 125 cm3125\text{ cm}^3 and 250 cm3250\text{ cm}^3, respectively. If the absolute temperature of the gas TWT_W at the point WW is such that nRTW=1nRT_W = 1 J (RR is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path XYXY is ____

Question
Correct Answer: 1.6

Detailed Solution

  1. For a monatomic gas, γ=5/3\gamma = 5/3 and Cp=52RC_p = \frac{5}{2}R.
  2. Path W→XW \to X is adiabatic. For an adiabatic process, TVγ−1=constantTV^{\gamma-1} = \text{constant}. TWVW2/3=TXVX2/3T_W V_W^{2/3} = T_X V_X^{2/3} TX=TW(VWVX)2/3=TW(64125)2/3=TW(45)2=0.64TWT_X = T_W \left(\frac{V_W}{V_X}\right)^{2/3} = T_W \left(\frac{64}{125}\right)^{2/3} = T_W \left(\frac{4}{5}\right)^2 = 0.64 T_W
  3. Path X→YX \to Y is isobaric. For an isobaric process, heat absorbed QXY=nCp(TY−TX)Q_{XY} = n C_p (T_Y - T_X).
  4. In an isobaric process, VT=constant\frac{V}{T} = \text{constant}, so TY=TX(VYVX)T_Y = T_X \left(\frac{V_Y}{V_X}\right). TY=(0.64TW)(250125)=1.28TWT_Y = (0.64 T_W) \left(\frac{250}{125}\right) = 1.28 T_W
  5. QXY=n(52R)(1.28TW−0.64TW)=52(nRTW)(0.64)Q_{XY} = n (\frac{5}{2}R) (1.28 T_W - 0.64 T_W) = \frac{5}{2} (nRT_W) (0.64).
  6. Given nRTW=1nRT_W = 1 J, we have: QXY=2.5×1×0.64=1.6 JQ_{XY} = 2.5 \times 1 \times 0.64 = 1.6 \text{ J} Final Answer: 1.6
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