Air expanding into chamber

  • Thread starter CAF123
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  • #26
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What do you think about this?

[itex]
dU=C_{V}dT\\\\
U=\int_{0}^{U}dU=\int_{T_{0}}^{T_{f}}C_{V}dT=C_{V}(T_{f}-T_{0})\\\\
U=\frac{3}{2}nRT\\\\
C_{V}(T_{f}-T_{0})=\frac{3}{2}nRT_{f}\\\\
\frac{C_{V}}{nR}(T_{f}-T_{0})=\frac{3}{2}T_{f}\\\\
\gamma=\frac{C_{P}}{C_{V}}=\frac{C_{V}+nR}{C_{V}}=1+\frac{nR}{C_{V}}\\\\
\frac{nR}{C_{V}}=\gamma-1\\\\
(\gamma-1)(T_{f}-T_{0})=\frac{3}{2}T_{f}\\\\
T_{f}=T_{0}\frac{1}{\frac{5}{2}-\frac{3}{2}\gamma}\\\\
\gamma=1.4\\\\
T_{0}=298K\\\\
T_{f}=745K
[/itex]
 
  • #27
21,636
4,913
What do you think about this?

[itex]
dU=C_{V}dT\\\\
U=\int_{0}^{U}dU=\int_{T_{0}}^{T_{f}}C_{V}dT=C_{V}(T_{f}-T_{0})\\\\
U=\frac{3}{2}nRT\\\\
C_{V}(T_{f}-T_{0})=\frac{3}{2}nRT_{f}\\\\
\frac{C_{V}}{nR}(T_{f}-T_{0})=\frac{3}{2}T_{f}\\\\
\gamma=\frac{C_{P}}{C_{V}}=\frac{C_{V}+nR}{C_{V}}=1+\frac{nR}{C_{V}}\\\\
\frac{nR}{C_{V}}=\gamma-1\\\\
(\gamma-1)(T_{f}-T_{0})=\frac{3}{2}T_{f}\\\\
T_{f}=T_{0}\frac{1}{\frac{5}{2}-\frac{3}{2}\gamma}\\\\
\gamma=1.4\\\\
T_{0}=298K\\\\
T_{f}=745K
[/itex]
Hi Tales. A hearty welcome to Physics Forms.

I hate to ruin your welcome, but there are numerous errors in your development.

1. For an ideal gas, with γ =1.4, U is not [itex]U=\frac{3}{2}nRT[/itex]. It is [itex]U=\frac{5}{2}nRT[/itex].
2. As a consequence of the this, for the notation that you are using, [itex]C_V=\frac{5}{2}nR[/itex]
3. Your third equation is not consistent with the first law. The term on the right hand side should be the work done in forcing the gas into the valve. This is just nRT0. There should be no 3/2, and the temperature should be T0.

Try redoing the problem with those corrections.

Chet
 

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