krootox217
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Homework Statement
I have the following task:
Homework Equations
The Attempt at a Solution
But I don't understand how to solve it. Can somebody help me?[/B]
Ideal gas volume work expression (adiabatic)
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I have the following task:
But I don't understand how to solve it. Can somebody help me?[/B]
Gold Member
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1) I'd guess they want you to apply the first law of thermodynamics, but I'm not sure.
2) Here you just have to do what the statement says (with a substitution considering ideal gas).
3) Based on the result in 2) you are able to proof that.
You need the ideal gas equation and the connectedness between R, cp and cv
2) Here you just have to do what the statement says (with a substitution considering ideal gas).
3) Based on the result in 2) you are able to proof that.
You need the ideal gas equation and the connectedness between R, cp and cv
vinicius0197
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First one: The differential form of the 1º law of thermodynamics is:
[itex]dU = dQ-dW[/itex]
Because it's an adiabatic process, there is no heat exchange, so Q = 0.
[itex]dU = -dW[/itex]
Internal energy is a function of state that depends solely on the quantity of gas - a number n of moles - and the absolute temperature. That can be written as:
[itex]dU = nC_vdT[/itex]
As for work:
[itex]dW = Fdx = PAdx = PdV[/itex]
Where F is force and Adx is the infinitesimal change in volume.
You can now easily see that:
[itex]nC_vdT = -PdV[/itex]
Second: First, remember that:
[itex]C_p = C_v + R[/itex]
For convenience, let's adopt that:
[itex]C_p/C_v = \gamma[/itex]
Using the famous equation, PV=nRT in the expression we derived in the first question:
[itex]nC_vdT = \frac{-nRT}{V}dV[/itex]
[itex]\frac{dT}{T}+\frac{R}{C_v}\frac{dV}{V} = 0[/itex]
And:
[itex]\frac{dT}{T}+(\gamma-1)\frac{dV}{V} = 0[/itex]
Integrating:
[itex]\int \frac{dT}{T} dT + (\gamma-1)\int \frac{dV}{V} = 0[/itex]
Which results in:
[itex]\ln (TV)^{(\gamma-1)} = cte[/itex]
And finally:
[itex](TV)^{(\gamma-1)} = cte[/itex]
For a initial state [itex](T_1, V_1)[/itex] and a final state [itex](T_2,V_2)[/itex]:
[itex]\frac{V_1}{V_2}^{(\gamma-1)} = \frac{T_2}{T_1}[/itex]
Third: Try to solve the last one for yourself. Now it's easy ;)
[itex]dU = dQ-dW[/itex]
Because it's an adiabatic process, there is no heat exchange, so Q = 0.
[itex]dU = -dW[/itex]
Internal energy is a function of state that depends solely on the quantity of gas - a number n of moles - and the absolute temperature. That can be written as:
[itex]dU = nC_vdT[/itex]
As for work:
[itex]dW = Fdx = PAdx = PdV[/itex]
Where F is force and Adx is the infinitesimal change in volume.
You can now easily see that:
[itex]nC_vdT = -PdV[/itex]
Second: First, remember that:
[itex]C_p = C_v + R[/itex]
For convenience, let's adopt that:
[itex]C_p/C_v = \gamma[/itex]
Using the famous equation, PV=nRT in the expression we derived in the first question:
[itex]nC_vdT = \frac{-nRT}{V}dV[/itex]
[itex]\frac{dT}{T}+\frac{R}{C_v}\frac{dV}{V} = 0[/itex]
And:
[itex]\frac{dT}{T}+(\gamma-1)\frac{dV}{V} = 0[/itex]
Integrating:
[itex]\int \frac{dT}{T} dT + (\gamma-1)\int \frac{dV}{V} = 0[/itex]
Which results in:
[itex]\ln (TV)^{(\gamma-1)} = cte[/itex]
And finally:
[itex](TV)^{(\gamma-1)} = cte[/itex]
For a initial state [itex](T_1, V_1)[/itex] and a final state [itex](T_2,V_2)[/itex]:
[itex]\frac{V_1}{V_2}^{(\gamma-1)} = \frac{T_2}{T_1}[/itex]
Third: Try to solve the last one for yourself. Now it's easy ;)
krootox217
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Thanks for the answers, I try to understand it and if I don't get everything, I ask again if that's ok :)