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C Secure Apps D My Fresno State https://wwwfli Welome, Daniei-8 MyMathLab Pearson Fip It Physics M Inbax -danielcallaha... M Inbox (5) -dicallahan 3406862&enrollmentio; 300623 1 2 3 4 The PV diagram below represents 2.86 mol of an ideal monatomic gas, The gas is initially at point A. The paths AD and BC represent isothermal changes. If the system is brought to point C along the path ADC, find the following: P atm 4.0 40 1) the initial temperature of the gas K Submit You currently have O submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 2) the final temperature of the gas K Submit You currently have 0 submissions for this question. Only 5 submission are allowed. You can make 5 more submissions for this question 3) the work done by the gasExplanation / Answer
(a)
Let, initial temperature = Ti
From ideal gas equation,
Pa*Va = nRTi
Take pressure in Pa and volume in m^3.
4*1.013*10^5 * 4.01*10^(-3) = 2.86*8.314*Ti
Ti = 68.33 K
(b)
Let final temperature = Tf
Pc*Vc = nRTf
1*1.013*10^5 * 20*10^(-3) = 2.86*8.314*Tf
Tf = 85.20 K
(c)
Given path AD and BC represents isothermic processes.
Volume at point D = Pa*Va / Pd
Vd = 4*4.01 / 1 = 16.04 L
Work done from A to D,
Wad = n R Ti * ln (Vd / Va)
Wad = 2.86*8.314*68.33 * ln (16.04 / 4.01)
Wad = 2252.38 J
Work done for path DC,
Wdc = 1*1.013*10^5 * (20*10^(-3) - 16.04*10^(-3))
Wdc = 401.1 J
Total work done by gas,
W = 2252.38 + 401.1
W = 2653.48 J
(d)
As Path AD and BC are isothermic,
interna energy From A to D = 0
lnternal energy for path DC,
Udc = 1.5*nR (Tc - Td) = 1.5*(Pc*Vc - Pd*Vd)
Udc = 1.5*(1.013*10^5*20*10^(-3) - 1.013*10^5*16.04*10^(-3))
Udc = 601.7 J
Heat absorbed by gas,
Q = U + W
Q = 601.7 + 2653.48
Q = 3255.18 J
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