Gas work -- Heating the gas in a cylinder with a weighted piston on top

In summary, the gas pressure inside the container changes as the piston rises. The work done by the gas on the piston is inversely proportional to the pressure. The temperature of the air inside the container is determined by the temperature of the gas at the time the piston reaches the top of the container.
  • #1
charlie05
128
6

Homework Statement


In the open cylindrical chamber is the piston of the total mass m. The initial air pressure inside the container is pa, the initial temperature T0. The initial height of the piston above the bottom h0. Now we start the gas supply heat to the moment when the piston reaches the height h above the bottom of the container .Air is ideal diatomic gas.

a / determine the final temperature T of the air inside the container

b / specify the total heat Q which air is received in the container

c / determine the efficiency of the piston stroke - ratio performed mechanical work and the total heat to the air inside the cylinder accepted

upload_2017-3-28_9-13-53.png

Homework Equations


F = mgΔh
W = F * ( h-h0) = p * S * ( h-h0) = p * ΔV
pV = nRT
paV0/T0 = p V/T

The Attempt at a Solution


[/B]
work necessary for the piston stroke of the mass m ...W = F*Δh = m*g*Δh = m*g* ( h-h0)
 
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  • #2
it is an adiabatic process? pVκ = konst.
 
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  • #3
charlie05 said:
it is an adiabatic process? pVκ = konst.
How can it be adiabatic if you are adding heat?
 
  • #4
charlie05 said:
work necessary for the piston stroke of the mass m ...W = F*Δh = m*g*Δh = m*g* ( h-h0)
Is there vacuum outside the container? If so, then you have calculated the work correctly.

From a force balance on the piston, how is mg related to pa and S? Does the gas pressure change? What is the work in terms of pa, S, and Δh? What is the initial volume in terms of h0 and S? What is the final volume in terms of h and S? From the ideal gas law, what is the final temperature in terms of h, h0, T0?
 
  • #5
yes, there occurs heat exchange, I understand...no vacuum, out is atmospheric pressure...

the initial volume V0 = S*h0
the final volume V = S*h

so answer a/ V/V0 = T/T0...T = T0*h/h0
 
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  • #6
b/ The first law of thermodynamics
Q = W + ΔU system

W = F*Δh = m*g*Δh = m*g* ( h-h0)

ΔU = Ek = 3/2 nRT...?
 
  • #7
charlie05 said:
b/ The first law of thermodynamics
Q = W + ΔU system

W = F*Δh = m*g*Δh = m*g* ( h-h0)

ΔU = Ek = 3/2 nRT...?
This is on the right track. But, if there is air outside then the work you calculated is not correct. From a force balance on the piston, $$P_aS=P_0S+mg$$where ##P_a## is the gas pressure and ##P_0## is the outside pressure. So the work the gas does on the piston is $$W=P_aS(h-h_0)=(P_0S+mg)(h-h_0)$$

Also, for a diatomic gas, the change in internal energy should be $$\Delta U=\frac{5}{2}nR(T-T_0)$$
 
  • #8
aha, thank you, now I see it...
c/ η = W/Q...?
 
  • #9
But I have a problem, I do not know molar amount n, I can not therefore calculate ΔU :-(
 
  • #10
charlie05 said:
But I have a problem, I do not know molar amount n, I can not therefore calculate ΔU :-(
Really? I think you have enough information to determine n. What do you think?
 
  • #11
Can I express it from the initial state?

pa * S * h0 = nRTo... n = (pa * V0)/R *T 0
 
  • #12
charlie05 said:
Can I express it from the initial state?

pa * S * h0 = nRTo... n = (pa * V0)/R *T 0
Yes.
 
  • #13
Q = W + ΔU = (p0S + mg ) ( h-h0 ) + 5/2 R ( T-T0) * ( (p0Sh0)/RT0 )
 
  • #14
charlie05 said:
Q = W + ΔU = (p0S + mg ) ( h-h0 ) + 5/2 R ( T-T0) * ( (p0Sh0)/RT0 )
The ##\Delta U## should have pa=p0, not p0. And, in that term, you should eliminate T and T0. Also, in the W term, you should use paS in place of (p0S + mg ).
 

Related to Gas work -- Heating the gas in a cylinder with a weighted piston on top

1. What is the purpose of heating gas in a cylinder with a weighted piston on top?

The purpose of this process is to increase the pressure and temperature of the gas inside the cylinder. This is commonly used in industries such as chemical production and gas storage.

2. How does heating the gas affect its pressure?

As the gas is heated, its molecules gain kinetic energy and move around more rapidly. This causes them to collide with the walls of the cylinder more frequently and with greater force, leading to an increase in pressure.

3. What is the role of the weighted piston in this process?

The weighted piston serves as a barrier to prevent the gas from expanding as it is heated. This allows the pressure inside the cylinder to increase, as the gas molecules have less space to move around.

4. What factors can affect the outcome of this process?

The temperature and pressure of the gas, as well as the weight and size of the piston, can all affect the outcome of heating gas in a cylinder. Other factors such as the volume of the cylinder and the type of gas being used can also play a role.

5. Is heating gas in a cylinder with a weighted piston a reversible process?

Yes, this process is reversible. If the temperature and pressure of the gas are lowered, the gas molecules will lose kinetic energy and the pressure will decrease. The weighted piston can then be removed to allow the gas to expand and return to its original state.

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