A 10.2-L canister of oxygen gas contains 58.6 mol of gas at a temperature of 25°C. What is the pressure of the gas? Looking to see how to set the problem up.

140.6 atm

Explanation:

Data Given:

Volume of Oxygen gas ( V )= 10.2 L

temperature of a gas ( T ) = 25 °C

Convert °C to K

T = °C + 273

T = 25 + 273 = 298 K

mol of oxygen gas (n) = 58.6

Pressure of a gas ( P ) =?

Solution:

Formula to be used

                  PV= nRT . . . . . . .(1)

As we have to find pressure so rearrange the above formula

                 P = nRT / V . . . . . . . . . . (2)

Where

n= number of moles

R is ideal gas constant

R = 0.0821 L atm mol⁻¹ K⁻¹

P = pressure

V = volume

T = temperature

Put values in equation 2

                 P = nRT / V

                 P = 58.6 mol x 0.0821 L atm mol⁻¹ K⁻¹ x 298 K / 10.2 L

                 P =  1434 L atm / 10.2 L

                 P = 140.6 atm

So the Pressure of oxygen gas (O₂) = 140.6 atm

Plasma state of the matter is the 4th statement and contains an ionized gas consisting of positive ions and free electrons in proportions resulting in more or less no overall electric charge, typically at low pressures !

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