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«Challenging» Thermodynamics Problem
Consider a cubical vessel of edge a, having a small hole in one of its walls. The total thermal resistance of the wall is [itex]\varphi[/itex] [itex]\mbox{At time} \ t=0[/itex], it contains air at atmospheric pressure [itex]p_a[/itex] and temperature [itex]\theta_0[/itex]The temperature of the surrounding is [itex]\theta_a ( > \theta_0 )[/itex] Find the amount of gas in moles in the vessel at time t. Take [itex]C_v = \frac{5R}{2}[/itex]
[tex]dQ=dW+dU[/tex]
[tex]dQ= \frac{i_{th}}{\varphi}[/tex]
[tex] PV=nR \theta \mbox{ideal gas eqn}[/tex]
I assumed pressure to be constant throughout the problem.
[tex]P=P_a[/tex]
Initially,
[tex]i_{th} = \frac{\theta_a - \theta_0}{\varphi}[/tex]
Now since volume and pressure both are constant,
PV=const.
or,
[itex]nRd\theta + R\theta dN = 0[/itex]
[itex]\frac{d\theta}{\theta} = -\frac{dn}{n}[/itex]
Now i try to apply first law, which gives,
[tex]\frac{\theta_a - \theta}{\varphi} dt = nC_vd\theta + \theta C_v n[/tex] where [itex]\theta\ is\ temperature \ at\ time \ t[/itex]
But since these rates are also varying, i have no idea how to continue. Specially if someone could throw light on the integration part.
Thanks for any assistance.
Homework Statement
Consider a cubical vessel of edge a, having a small hole in one of its walls. The total thermal resistance of the wall is [itex]\varphi[/itex] [itex]\mbox{At time} \ t=0[/itex], it contains air at atmospheric pressure [itex]p_a[/itex] and temperature [itex]\theta_0[/itex]The temperature of the surrounding is [itex]\theta_a ( > \theta_0 )[/itex] Find the amount of gas in moles in the vessel at time t. Take [itex]C_v = \frac{5R}{2}[/itex]
Homework Equations
[tex]dQ=dW+dU[/tex]
[tex]dQ= \frac{i_{th}}{\varphi}[/tex]
[tex] PV=nR \theta \mbox{ideal gas eqn}[/tex]
The Attempt at a Solution
I assumed pressure to be constant throughout the problem.
[tex]P=P_a[/tex]
Initially,
[tex]i_{th} = \frac{\theta_a - \theta_0}{\varphi}[/tex]
Now since volume and pressure both are constant,
PV=const.
or,
[itex]nRd\theta + R\theta dN = 0[/itex]
[itex]\frac{d\theta}{\theta} = -\frac{dn}{n}[/itex]
Now i try to apply first law, which gives,
[tex]\frac{\theta_a - \theta}{\varphi} dt = nC_vd\theta + \theta C_v n[/tex] where [itex]\theta\ is\ temperature \ at\ time \ t[/itex]
But since these rates are also varying, i have no idea how to continue. Specially if someone could throw light on the integration part.
Thanks for any assistance.
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