- #1

- 26

- 0

I am overwhelmed by this exercise. Normally I could make the equation (S is the integral sign) S Pdt = S nRdT (n=pV/RT), but now more air is flowing in, all the time warming the room and making the n change. How the hell does this work?

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- Thread starter Kelju Ivan
- Start date

- #1

- 26

- 0

I am overwhelmed by this exercise. Normally I could make the equation (S is the integral sign) S Pdt = S nRdT (n=pV/RT), but now more air is flowing in, all the time warming the room and making the n change. How the hell does this work?

- #2

- 38

- 0

[tex]p_{0}V_{0} = N(t)k_{b}T(t) \Rightarrow N(t) = \frac{p_{0}V_{0}}{k_{b}T(t)}[/tex]

Next, we can right the power as

[tex]P = \frac{dQ}{dt} = C_{V}\frac{dT}{dt}[/tex]

It says that you have a diatomic gas. We know that the heat capacity of a diatomic gas is

[tex]C_{V} = \frac{5}{2}Nk_{b}[/tex]

Substituting for the heat capacity and the number of particles,

[tex]P = \frac{5p_{0}V_{0}}{2k_{b}T(t)}\frac{dT}{dt}[/tex]

Thus,

[tex]P\Delta t = \frac{5p_{0}V_{0}}{2k_{b}} ln \frac{T_{final}}{T_{initial}}[/tex]

- #3

- 26

- 0

In our version we had [tex] C_p=\frac{7}{2}R [/tex]. I'm not really sure when I should use 5/2 and when 7/2, since both are valid for [tex]C_p[/tex].

[tex]t=Q/P[/tex] where

[tex] Q=nC_p\Delta T [/tex] and [tex] n=\frac{pV}{RT}[/tex]

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