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

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.