Consider a gas in a cylinder at room temperature (T = 293 K), with a volume of 0
ID: 1472890 • Letter: C
Question
Consider a gas in a cylinder at room temperature (T = 293 K), with a volume of 0.125 m3. The gas is confined by a piston with a weight of 150 N and an area of 0.75 m2. The pressure above the piston is atmospheric pressure.
A.) What is the absolute pressure of the gas?
B.) How many moles of the gas are there? How many individual gas particles are there?
C.) The gas is heated, expanding it and moving the piston up. If the volume occupied by the gas triples, how much work is done on the gas if the pressure is held constant during the process?
D.) If the walls of the cylinder and the piston itself are good thermal insulators, no heat (Q) flows into or out of the system. Use the first law of thermodynamics and your answer from part ‘b’ to calculate the change in temperature of the gas in both Kelvin and degrees Celsius.
Explanation / Answer
A) P_absolute = Po + W/A
= 1.013*10^5 + 150/0.75
= 1.015*10^5 pa
B) use Ideal gas equation, P*V = n*R*T
n = P*V/(R*T)
= 1.015*10^5*0.125/(8.314*293)
= 5.208 mol
no of molecules, N = n*NA
= 5.308*6.023*10^23
= 3.197*10^24 molecules
C) Workdone = P*(V2-V1)
= P*(3*V1 - V1)
= 1.015*10^5*(3*0.125 - 0.125)
= 25375 J
D) when dQ = 0
from First law of thermodynamics,
dQ = dU + dW
=> dU = -dW
= -25375 J
we know, dU = (3/2)*n*R*dT
==> dT = dU/(1.5*n*R)
= -25375/(1.5*5.308*8.314)
= -383.3 K
negative sign indicates decrease in temperature.
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