I am stuck on this question. I took a screen shot of the question, please provie
ID: 2270990 • Letter: I
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I am stuck on this question. I took a screen shot of the question, please provied the forumula you used as well. Thanks
At point D in the figure below, the pressure and temperature of 2.00 mol of an ideal diatomic gas are 2.00 atm and 360 K, respectively. The volume of the gas at point B on the PV diagram is three times that at point D and its pressure is twice that at point C. Paths AB and CD represent isothermal processes. The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas and the heat absorbed by the gas along each portion of the cycle. Wby gas,tot = kJ QD rightarrow A = kJ QA rightarrow B = kJ QB rightarrow C = kJ QC rightarrow D = kJExplanation / Answer
At D, pressure PD = 2 atm , Temperature TD = 360 K , number of moles = 2 mol.Let VD be the Volume of gas at D .
By ideal gas law, we have P*V = n*R*T
so, 2* VD = 2* 0.0821 * 360
VD = 29.556 liter.
At point B , we have VB = 3*VD = 3* 29.556 = 88.668 liters.
Pressure PB = 2*PC
As CD and AB are isothermal , we have TC = TD ; TA = TB.
so, by ideal gas law,
(PB)*(VB) = n*R*(TB)
(2*PC) * (88.668) = 2 * 0.0821 * TB
so, PC = 0.0009259 TB
By ideal gas law,
(PC)*(VC) = n*R*TC
0.0009259 TB * (VC) = 2*0.0821 * TC
we have VC = VB = 88.668 liter, TC = TD = 360 K
so, 0.0009259 TB * (88.668) = 2*0.0821 * 360
TB = 720.02 K
so, TA = TB = 720.02 K , TC= TD = 360 K
VA = VD = 29.556 liter. VB = VC = 88.668 litre.
PD = 2 atm
we have (PD) * (VD) / TD = (PC) * ( VC)/ TC
2 * 29.556 / 360 = PC * 88.668 / 360
so, PC = 0.6666 atm
we have PB = 2*PC = 2* 0.6666 = 1.333 atm.
we have (PD) * (VD) / TD = (PA) * ( VA)/ TA
2 * 29.556 / 360 = PA * 29.556 / 720.02
PA = 4 atm.
so, PA = 4 atm , VA = 29.556 liter , TA = 720.02 K
PB = 1.333 atm. , VB = 88.668 litre. , TB = 720.02 K
PC = 0.6666 atm , VC = 88.668 litre , TC = 360 K
PD = 2 atm , VD = 29.556 liter , TD = 360 K
i) From D to A , Isochoric process
work done from D to A= PD * ( VD - VA) = 0
Heat supplied = change in internal energy = dU = n* Cv * ( TA- TD)
cv for diatomic ideal gas = 2.5 * R = 2.5 * 0.0821 = 0.20525
so, dU = 2* 0.20525 * ( 720.2 - 360 ) = 147.8621 L.atm = 14982.127 Joules
so, Heat supplied =Q D -> A = 14982.127 Joules = 14.982 KJ
ii) From A to B , change in internal energy = 0
work done during isothermal process = n*R*TA * ln ( VB / VA)
work done = 2* 0.0821 * 720.02 * ln ( 88.668 / 29.556) = 129.8859 L. atm = 13160.688 J
work done during A-> B = 13160.688 KJ = 13.160 KJ
so heat given during A-> B = Q A-> B = 13.160 KJ
iii) B to C , isochoric process
work done from B to C = PB * ( VB - VC) = 0
Heat supplied = change in internal energy = dU = n* Cv * ( TC- TB)
cv for diatomic ideal gas = 2.5 * R = 2.5 * 0.0821 = 0.20525
so, dU = 2* 0.20525 * ( 360 - 720.2 ) = - 147.8621 L.atm = -14982.127 Joules
so, Heat supplied =Q B -> C = -14982.127 Joules = -14.982 KJ
iv) C to D ,change in internal energy = 0
work done during isothermal process = n*R*TC * ln ( VD / VC)
work done = 2* 0.0821 * 360 * ln ( 29.556/88.668 ) = -64.941 L. atm = -6580.146 J
work done during C-> D = -6580.146 KJ =-6.580 KJ
so heat given during C-> D = Q C-> D = -6.580 KJ
Q D -> A = 14982.127 Joules = 14.982 KJ
Q A-> B = 13.160 KJ
Q B -> C = -14982.127 Joules = -14.982 KJ
Q C-> D = -6.580 KJ
Total heat absorbed by the gas = (Q D -> A) + (Q A-> B ) + (QB -> C ) +(Q C-> D)
Total heat absorbed by the gas = 14.982 + 13.16 -14.982 -6.58 = 6.58 KJ
so work done during the process by gas = Total heat absorbed by the gas = 6.58 KJ
W by gas = 6.58 KJ
Q D -> A = 14982.127 Joules = 14.982 KJ
Q A-> B = 13.160 KJ
Q B -> C = -14982.127 Joules = -14.982 KJ
Q C-> D = -6.580 KJ
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