720 grams of an ideal monatomic gas at pressure 3 atm temperature 300 K occupies

Question

720 grams of an ideal monatomic gas at pressure 3 atm temperature 300 K occupies 50 L. Find the number of moles, the total translational kinetic energy, and the rms speed of the molecules. The gas is heated at constant volume to pressure 4 atm. Find the final temperature and the heat added. The gas Is returned to state ft and compressed to volume 30 L at constant pressure (3 atm), to state C. Find the final temperature, the work, the internal energy change, and the heat transfer (in/out rightarrow +/-): The gas is again returned to state A and compressed to 30 L, this time isothermally-at constant temperature (300 K). Find the final pressure. the work, the internal energy change, and the heat transfer. Draw the graph for this process in the diagram above, and indicate the final stale D. The gas is again returned to state A and compressed to 30 L, this time adiabatically (no heat transfer). The final pressure is 7 atm. Check that this is right for a monatomic gas: Find the final temperature, the internal energy change, and the work.

Explanation / Answer

PA = 3 atm - 3*10^5 pa

VA = 50 L = 50*10^-3 m^3

TA = 300 K

mass m = 0.72 kg

part (a)

number of moles n = PA*VA/(R*TA) = (3*10^5*50*10^-3)/(8.314*300) = 6.014 moles

KTR = (3/2)*n*R*T = (3/2)*6.014*8.314*300 = 22.5 KJ

(1/2)*m*v^2 = KE

v = sqrt(2*KE/m) = sqrt(2*22500/0.72) = 250 m/s

--------------------------

paart (b)

at constant volume

TB/TA = PB/PA

TB/300 = 4/3

TB = 400 K

QA->B = n*Cv*dT = n*(3/2)*R*(TB - TA)

QA-->B = 6.014*(3/2)*8.314*(400-300) = 7.5 KJ

========================================

part(c)

at constant pressure

Tc/TA = Vc/VA

Tc/300 = 30/50

Tc = 180 K

WA---> c = PA*dV = PA*(Vc-VA) = 3*10^5*(30-50)*10^-3 = -6 KJ

d_EA-->c = n*C*vdT = n*(3/2)*R*(Tc-TA) = 6.014*(3/2)*8.314*(180-300) = -9 KJ

QA--c = -15 KJ

===========================================

at constant temperature

PD*VD = PA*VA

PD*30 = 3*50

PD = 5 atm

WA--B = 2.303*n*R*TA*log(pA/PD) = 2.303*6.014*8.314*300*log(3/5) = -7.6638 kJ

QA--D = WA--D = -7.6638 kJ

-----------------

for adiabatic process

gamma r = 5/3

TE*VE^(r-1) = TA*VA^(r-1)

TE*30^(5/3-1) = 300*50^(5/3-1)

TE = 421.7 K

dE = n*Cv*dT = 6.014*(3/2)*8.314*(421.7-300) = 9.13 kJ

WA-E = -dE = -9.13 kJ

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