A cylinder of oxygen gas contains 2.80 mol of 02. If the volume of the cylinder is 8.58 L, what is the pressure of the 02 if the gas temperature is 294 K.

pressure of the oxygen gas is 7.88atm

Explanation:

Moles of oxygen in cylinder  =  2.80 mol

volume=8.58L

temperature=294K

molar gas constant(R)= 0.0821 L atm / (mol K)

The pressure is calculated using ideal gas law:

PV = nRT

making Pressure subject of the formula

P = nRT /V

  = 2.80× 0.0821 ×294  /8.58 L

P = 7.88 atm

7.88 atm

Explanation:

This question is a very straight forward Question, we have given almost all the parameters needed to solve the question. The parameters given in the question are: Number of moles of oxygen gas = 2.80 mol, the volume of the cylinder= 8.58 L, pressure of the 02 = ???( not given) and the gas temperature = 294 K.

We are going to use the formula for the ideal gas law below to solve this question:

PV = nRT.

Where P = pressure, V = volume , n= number of moles, R= gas constant and T= temperature.

nRT/ V= P.

P= 2.80 × 0.0821 × 294/ 8.58.

P= 67.58472÷ 8.58.

P= 7.87700699300699300.

P= 7.88 atm.

the answer is b. dilution