# Does Gay Lusaac's law work both ways

1) With $n,V$ constant, increasing $T$ results in an increase in $P$ : $$P = k_1T$$

2) With $n,V$ constant, increasing $P$ results in an increase in $T$ : $$T = k_2P$$

We can do the first process by simply heating the confined gas.

However I'm not able to think of a way to do the second process.
How can I make "increasing $P$" the cause of "increasing $T$" ? (with out changing volume or adding more matter)

Basically there are only 3 ways of increasing the pressure.

1. By decreasing the volume of the gas (i.e. decreasing V)

2. By adding more amount of gas (i.e. increasing number of moles "n")

3. By increasing the temperature of the gas (i.e. increasing T)

Now your question being, would increase in pressure "P" keeping "V" and "n" constant result in the rise in temperature "T". So at the first place you need to increase the pressure.

By keeping V and n constant point 1 and point 2 discussed above is ruled out.

Now you want to observe T from changing P that means T, i.e., temperature over here cannot be used to increase P because you want to see how temperature varies with pressure, i.e. temperature here is to be observed.

So temperature cannot be used here to rise pressure. How would you increase pressure now? Out of the three factors which could have increased the pressure, two of the factors are constant and the last factor, i.e., temperature factor is to be observed and hence cannot be used to rise the pressure. So how will you now increase the pressure? It's not possible to increase the pressure without using any of the above factors.

As you are not able to increase the pressure in the first place, then how are you expecting to see the variation in temperature?

• Awesome! so there is no way to change pressure with out touching one of $n,P,T$. Thanks a lot for the clear explanation (: – Hiiii Mar 18 '17 at 8:52