Please try to solve the question with proving the work: Suppose that 248 moles o
ID: 1843328 • Letter: P
Question
Please try to solve the question with proving the work:
Suppose that 248 moles of a monatomic ideal gas is initially contained in a piston with a volume of 0.73 m3 at a temperature of 522 K. The piston is connected to a hot reservoir with a temperature of 1296 K and a cold reservoir with a temperature of 522 K. The gas undergoes a quasi-static Stirling cycle with the following steps: 1. The temperature of the gas is increased to 1296 K while maintaining a constant volume. 2. The volume of the gas is increased to 2.96 m3 while maintaining a constant temperature. 3. The temperature of the gas is decreased to 522 K while maintaining a constant volume. 4. The volume of the gas is decreased to 0.73 m3 while maintaining a constant temperature t may help you to recall that Cv 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 x 23 times the number of moles of the gas. 1) What is the pressure of the gas under its initial conditions? Pa Submit 2) How much energy is transferred into the gas from the hot reservoir? J Submit 3) How much energy is transferred out of the gas into the cold reservoir? J SubmitExplanation / Answer
1)
Initial volume V1 = 0.73 m3
Initial temperature T1 = 522 K
For ideal gas, PV = nRT......where n = number of moles, R is universal gas constant = 8.314 J/mol-K
P1*0.73 = 248*8.314*522
P1 = 1474379.7 Pa
2)
Using first law of thermodynamics, Q - W = U2 - U1 where Q is heat treansfer, W is work done, U is internal energy
1-2 is a constant volume process, so W = 0
Hence, Q = U2 - U1 = n*Cv*(T2 - T1)
Q = 248*12.47*(1296 - 522)
Q12 = 2393641.44 J
2-3 is a constant temperature process. Hence, U2 = U3
Q - W = 0
Q23 = W = nRT ln (V3 / V2)
= 248*8.314*1296 ln (2.96 / 0.73)
= 3740793.4 J
Total energy transfer = Q12 + Q23 = 6134434.8 J
3)
Similarly, Q34 = W34 + (U4 - U3)
= 0 + 248*12.47*(522 - 1296)
= - 2393641.44 J
Similarly, Q41 = W41 = nRT ln (V1 / V4)
= 248*8.314 * 522 * ln (0.73 / 2.96)
= -1506708.44 J
Net energy out = Q34 + Q41
= -3900349.9 J
4)
Net work done = Net energy transfer
= Energy in + Energy out
= 6134434.8 - 3900349.9
= 2234084.9 J
5)
Efficiency = Net work / Energy in
= 2234084.9 / 6134434.8
= 0.364
6)
Carnot Efficiency = 1 - T_low / T_high
= 1 - 522 / 1296
= 0.597