A gas has a pressure ot 34200 mmHg at an unknown temperature. When the pressure inside of the tank is reduced to l 1400 mmHg and has a temperature of 35 C. What was the original temperature of the gas inside of the tank?

1 Answer
Nov 3, 2016

The quoted units of pressure are absurd. From where are you getting them?

Explanation:

1*atm1atm will support a column of mercury 760760 mmmm high. You can use a mercury column to measure a reduced pressure, or you could use a mercury bubbler as a control to release excess gas, but you would always be aware of the possibility that if you break your glassware, or have some other accident, events which are all too possible in a laboratory, you will get mercury all over your bench, where the metal will inhabit every nook and cranny. This will be a major clean up job, which a contract cleaner would not touch.

So, to your problem, we have a starting pressure of,

"34200 mm Hg"/("760 mm Hg "atm^-1)34200 mm Hg760 mm Hg atm1 == 45*atm45atm, P_1P1.

And an end pressure of,

"1400 mm Hg"/("760 mm Hg "atm^-1)1400 mm Hg760 mm Hg atm1 == 1.84*atm1.84atm, P_2P2.

Using Charles' Law, (P_1)/T_1=(P_2)/T_2P1T1=P2T2

So T_1=(P_1)/(P_2)xxT_2T1=P1P2×T2 == (45*atm)/(1.84*atm)xx308*K45atm1.84atm×308K, which is a rather high temperature.

I reiterate that the question is highly suspect, and the units are physically impossible.