What pressure is argon at room temperature?

Answer 1

The given information is not sufficient to answer this question.

You can use the Ideal Gas Law to find out though, expressed mathematically as:

PV=nRT

Where:

P=Pressure (in mmHg)

V=Volume (in Liters)

n=number of moles of gas

R=62.36367 L·mmHg·K−1·mol−1 (Ignore the jargon at the end just know that solving the equation for P will give an answer in the unit mmHg,)

T= Temperature (in Kelvin) (room temperature in Kelivin is 293 K)

You would already need to know V and n to begin with in order to be able to do this equation, however for the sake of example (exactly) one liter and .0094 mole of Argon would be:

P(1)=(.0094)(62.36367)(293)

P= 170 mmHg

One mole or 39.948 grams of Argon would be at a pressure of

P(1)=(1)(62.36367)(293)

P= 18300 mmHg which is 24 times the pressure of Earth's atmosphere.