gay-lussac gas law

Question

If a tank of gas at standard pressure and 932.6*C is increased to 810. mmHg, what is the new temperature? Don't forget to convert to K.

(hint: What's the standard pressure for this unit? This will be what you type in the box!) -that's what my chemistry teacher put I have no clue why

Report ANS in the correct sig-figs (only use the data from the problem to calculate the sig figs! Ignore the conversion factors)

Your Answer:

Answer 1

The equation you will be using to answer this question is:

P1/T1=P2/T2

In order to use this equation your temperature needs to be in kelvin and your pressure can be in any unit as long as you stay consistent, in this case we will use atm as that is probably the "standard" pressure to which your teacher is referring

Standard pressure is 1atm or 760 mmHg

So P1 is 1atm

T1= 932.6 +273.15 = 1205.75 K

P2= 810 X (1atm/760mmHg) = 1.06578

T2 is what we are solving for

Plug all the numbers in to get your answer!