# 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?

Nov 3, 2016

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

#### Explanation:

$1 \cdot a t m$ will support a column of mercury $760$ $m m$ 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) $=$ $45 \cdot a t m$, ${P}_{1}$.

And an end pressure of,

"1400 mm Hg"/("760 mm Hg "atm^-1) $=$ $1.84 \cdot a t m$, ${P}_{2}$.

Using Charles' Law, $\frac{{P}_{1}}{T} _ 1 = \frac{{P}_{2}}{T} _ 2$

So ${T}_{1} = \frac{{P}_{1}}{{P}_{2}} \times {T}_{2}$ $=$ $\frac{45 \cdot a t m}{1.84 \cdot a t m} \times 308 \cdot K$, which is a rather high temperature.

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