Suppose that 1.6 moles of a monatomic ideal gas (atomic mass = 2.2 × 10 -27 kg)
ID: 1508605 • Letter: S
Question
Suppose that 1.6 moles of a monatomic ideal gas (atomic mass = 2.2 × 10-27 kg) are heated from 300 K to 500 K at a constant volume of 0.16 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1.
How much energy is transferred by heating during this process?
2.
How much energy is transferred by heating during this process?
3.
What is the pressure of the gas once the final temperature has been reached?
4.
What is the average speed of a gas molecule after the final temperature has been reached?
5.
The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?
6.
How much work was done on or by the gas during the constant-pressure process?
Explanation / Answer
1 and 2) At constant volume
Energy transferred = nCvT
n = 1.6
Energy transferred = 1.6*12.47*(500-300) = 3990.4 J
3) At constant volume
P1/T1 = P2/T2
T1 = 300 K
V1 = 0.16 m^3
P1V1 = nRT1
P1 = nRT1 / V1 = 1.6*0.082*300 / 0.16 = 246 atm
246/300 = P2/500
P2 = 410 atm
4) Average speed = sqrt(3RT/M)
No of molecules = n*Avogadro no = 1.6*6.022*10^23 = 9.63*10^23
M = Molar mass = 9.63*10^23 * 2.2 × 10^-27 kg = 2.12*10^-3 kg
Average speed = sqrt(3*0.082*500/2.12*10^-3) = 240.87 m/s
5) during the constant-pressure process
Energy transferred = nCpT = 1.6*20.79*200 = 6652.8 J
6) Work done = P1V = P1(V2 - V1)
At constant pressure
V1/T1 = V2/T2
0.16/300 = V2/500
V2 = 0.267 m^3
Work done = P1V = P1(V2 - V1) = 246*(0.267-0.16) = 26.32 J
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