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

the correct answer would be d. hope this !