6.0 liters of a monoatomic ideal gas, originally at 400K and a pressure of 3.0 a
ID: 1401304 • Letter: 6
Question
6.0 liters of a monoatomic ideal gas, originally at 400K and a pressure of 3.0 atm (called state 1) undergoes the following processes: 1 -> 2 isothermal expansion to V2 = 4V1 2 -> 3 isobaric compression 3 -> 1 adiabatic compression to its original state (a) Find the pressure, volume, and temperature of the gas in state 2. (b) Find the pressure, volume, and temperature of the gas in state 3. (c) How many moles of gas are there? (d) How much work is done by the gas during each of the three processes? (e) How much heat flows into the gas in each of the three processes?
Explanation / Answer
1 ---> 2 Isothermal, so temperature is constant
P1*V1 = P2*V2
P2 = P1*V1/V2
P2 = (3*V1)/4V1 = 3/4 = 0.75 atm <<<------------answer
V2 = 4V1 = 4*6 = 24 liters <<<------------answer
T2 = T1 = 400 K <<<------------answer
3 --->>1 adiabatic compression
V1 = 6
P3 = 0.75 atm
P1 = 3 atm
from adiabatic proess equation
PV^gamma = constant
P3*V3^gamma = P1*V1^gamma
gamma = ratio of specific heat = 1.67
0.75*V3^1.67 = 3*6^1.67
v3 = 13.762 liters
2 ------>3 isobaric process, pressure is constant
v3 = 13.762 liters <<<------------answer
P3 = P2 = 0.75 atm <<<------------answer
V3 = 13.762 liters
T3/T2 = V3/V2
T3 = (T2*V3)/V2
T3 = (400*13.762)/24 = 229.4 K <<<------------answer
part(c)
n = P1*V1/R*T1
n = (3*1.013*10^5*6*10^-3)/(8.314*400) = 0.548 <-------answer
part(d)
W1-->2 = 2.303*n*R*T1*log(V2/V1) = 2.303*0.548*8.314*400*log4
W 1-->2 = 2526.8 J <<<------answer
W 2-->3 = P*dV = P2*(V3-V2) = 0.75*1.013*10^5*(13.762-24)*10^-3
W2--3 = -777.832 J <<<---answer
W3--1 = (P1*V1 - P3*V3)/(1 - gamma)
W 3--1 = ((3*1.013*10^5*6*10^-3)-(0.75*1.013*10^5*13.762*10^-3))/(1-1.67)
W3--1 = -1160.9 J
(e)
from 1 -->2
temperature is conatnt
change in internal energy dU = 0
from first law of thermodynamice dQ = dU + dW
Q1--2 = W1--2 = 2526.8 J
from 2 --- >3
dU = n*R*dT/(gamma-1)
dU = (0.548*8.314*(229.4-400))/(1.67-1) = -1160.9 J
dQ 2--3 = dU + dW = 1937.93 J
from 3 -->1 adiabatic process
dQ3--1 = 0
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