At what temperature will 0.606 mole of oxygen exert a pressure of 2.4 atm

in a 5.12 L cylinder?

At a temperature of 251.15 K

Explanation:

To calculate this temperature, we make use of the ideal gas equation

Mathematically;

PV = nRT

where P is the pressure given as 2.4atm

V is the volume which is the volume of the cylinder at 5.2L

n is the number of moles which is 0.606 according to the question

R is the molar gas constant which is 0.082 L atm K^-1 mol^-1

T is the temperature which we are looking for according to the question

Rearranging the equation, we have;

T = PV/nR

Plugging the values;

T = (2.4 * 5.2) / (0.606 * 0.082)

T = 12.48/0.049692

T = 251.15 K