# Throttling a Gas Cylinder

by jonathan123
Tags: cylinder, throttling
 P: 32 Your question needs more detail. It is not apparent which quantities are being held fixed here. If it is adiabatic expansion then the volume V is obviously changing. There is no thermodynamic process (except diffusion) where P,V, and T all remain constant. I'll assume you have an insulated gas expanding due to some valve being opened: The relation you need to use is $P_1{V_1}^\gamma=P_2{V_2}^\gamma$, assuming the gas is ideal. The derivation of this relation can be found at http://en.wikipedia.org/wiki/Isentropic_process
 P: 32 Throttling a Gas Cylinder Ok, so it IS adiabatic expansion. The formula I have written in my last post relates P and V at different times. However, you should also remember that the equation of state PV=NRT is always valid for an ideal gas. So you can easily transform equation $P_1{V_1}^\gamma=P_2{V_2}^\gamma$ in terms of P and T as $\frac{{P_1}^{\gamma-1}}{{T_1}^\gamma}=\frac{{P_2}^{\gamma-1}}{{T_2}^\gamma}$. And you are done.
 P: 32 $\frac{P^{\gamma-1}}{T^\gamma}$ is a conserved quantity for the adiabatic process. No matter what instance of time you take during the process, the instantaneous pressure and temperature will yield a constant value of $\frac{P(t)^{\gamma-1}}{T(t)^\gamma} = \frac{{P_1}^{\gamma-1}}{{T_1}^\gamma}$. Also we know $P(t)V(t)=NRT(t)$. You know values of V(t), $P_1$ and $T_1$. You have two equations in two unknowns. Solve for P(t) and T(t).