# Gases and physical chemistry

PV=nRT. Hence, T=PV/nR.Let's consider a gas containing in a
bulb in a way that its pressure,volume and number of moles of gas can be changed.So if I want to attain a certain temperature for n moles of a gas in the bulb,I can adjust the PV value in the equation so as to fit to the value of T.Theoretically this can be done with above equation.But is this experimentally possible?

• Yes. Related: physics.stackexchange.com/questions/17948/… Feb 4, 2017 at 19:08
• Nonsense. You cannot "change" the pressure actively, except by changing n,V or T. An ideal gas especially does not change it's temperature when compressed. The Joule-Thomson coefficient of an ideal gas is zero.
– Karl
Feb 4, 2017 at 19:50
• @Karl Wouldn't that violate conservation of energy though? physics.stackexchange.com/questions/136408/… Feb 4, 2017 at 21:14
• If the system is adiabatic and you apply work to it, that energy has to go somewhere. If I recall, ideal gases do not go through phase changes. Feb 4, 2017 at 21:17
• @QuantumAMERICCINO You increase the pressure of an ideal gas by (anti-)proportionally lowering the volume. Exactly antiproportional. No temperature change occurs, because otherwise the result would not be an ideal gas.
– Karl
Feb 4, 2017 at 21:39

I will try to summarize what Karl is saying about the ideal gases. Suppose that you measure the temperature of a gas contained in a canister. Suppose that you measure the temperature $T_i$
Using your equation that you derived from $PV=nRT$, we should have $$T_i=\frac{P_iV_i}{nR}$$ Suppose that we then apply work to the canister. Since it is an ideal gas the only way that we can change its energy is through a change in volume ,pressure, or number of moles. So that is what we are going to do. Suppose that we compress the canister so that the pressure of the gas changes. Thus, our new canister will have a new pressure $P_f$ and a new volume $V_f$. This volume $V_f$, however, is related to $V_i$ through the ideal gas equation $$P_iV_i=P_fV_f$$ Hence we have that $$V_f=\frac{P_iV_i}{P_f}$$ Lets substitute this equation to find the new temperature of our compressed canister:
$$T_{compressed}=\frac{P_f\frac{P_iV_i}{P_f}}{nR}$$ Which gives us the initial equation. Hence the temperature of the ideal doesn't change when it is compressed.