Question

In a storage area of a hospital where the temperature has reached 55 °C, the pressure of oxygen gas in a 15.0-L steel cylinder is 965 Torr.What is the final pressure, in millimeters of mercury, when the temperature of the oxygen gas drops to 24 °C, and the volume and the amount of the gas do not change?

Answer 1

`873.80\text{ mmHg}`

Step-by-step explanation:

Here, we want to get the final pressure

From the pressure law, volume and temperature are directly proportinal

Mathematically:

`(P_1)/(T_1)\text{ = }(P_2)/(T_2)`

Where:

P1 is the initial pressure which is 965 torr

P2 is ?

T1 is the initial temperature which we convert to Kelvin by adding 273 K(55 + 273 = 328 k

T2 is the final temperature which is 24 + 273 = 297 K

Substituting the values, we have it that:

`\begin{gathered} (965)/(328)\text{ = }(P_2)/(297) \\ \\ P_2\text{ = }(297*965)/(328)\text{ = 873.80 torr} \end{gathered}`

Now, we convert this to mmHg

Mathematically, 1 torr = 1 mmHg

We have the final pressure as 873.80 mmHg