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

[charlie05]

Homework Statement

In the open cylindrical chamber is the piston of the total mass m. The initial air pressure inside the container is p_a, 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

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)

 

[charlie05]

it is an adiabatic process? pV^κ = konst.

 

[Chestermiller]

it is an adiabatic process? pV^κ = konst.

How can it be adiabatic if you are adding heat?

 

[Chestermiller]

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 p_a and S? Does the gas pressure change? What is the work in terms of p_a, 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?

 

[charlie05]

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

 

[charlie05]

b/ The first law of thermodynamics
Q = W + ΔU _system

W = F*Δh = m*g*Δh = m*g* ( h-h₀)

ΔU = E_k = 3/2 nRT...?

 

[Chestermiller]

b/ The first law of thermodynamics
Q = W + ΔU _system

W = F*Δh = m*g*Δh = m*g* ( h-h₀)

ΔU = E_k = 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 $$\Delta U=\frac{5}{2}nR(T-T_0)$$

 

[charlie05]

aha, thank you, now I see it...
c/ η = W/Q...?

 

[charlie05]

But I have a problem, I do not know molar amount n, I can not therefore calculate ΔU :-(

 

[Chestermiller]

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?

 

[charlie05]

Can I express it from the initial state?

p_a * S * h₀ = nRT_o... n = (p_a * V₀)/R *T ₀

 

[Chestermiller]

Can I express it from the initial state?

p_a * S * h₀ = nRT_o... n = (p_a * V₀)/R *T ₀

Yes.

 

[charlie05]

Q = W + ΔU = (p₀S + mg ) ( h-h₀ ) + 5/2 R ( T-T0) * ( (p₀Sh₀)/RT₀ )

 

[Chestermiller]

Q = W + ΔU = (p₀S + mg ) ( h-h₀ ) + 5/2 R ( T-T0) * ( (p₀Sh₀)/RT₀ )

The $\Delta U$ should have p_a=p₀, not p₀. And, in that term, you should eliminate T and T₀. Also, in the W term, you should use p_aS in place of (p₀S + mg ).