Question

How many moles of gas must be forced into a 3.8 L ball to give it...

How many moles of gas must be forced into a 3.8 L ball to give it a gauge pressure of 9.6 psi at 27 ∘C? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is 14.9 psi so that the total pressure in the ball is 24.5 psi .

No information is given; assume the gas to behave ideally and apply the ideal gas law. Given the pressure of the gas is 24.5 psi (the gas fills up the balls and hence, the gas exerts the same pressure) and 1 atm = 14.6959 psi, we find the pressure in atmospheres as

P = (24.5 psi)*(1 atm/14.6959 psi) = 1.6671 atm.

The volume of the gas is V = 3.8 L and the temperature of the gas is T = 27°C = (27 + 273.15) K = 300.15 K.

Use the ideal gas law, P*V = n*R*T where n = number of moles of the gas; therefore,

(1.6671 atm)*(3.8 L) = n*(0.082 L-atm/mol.K)*(300.15 K)

====> n = (1.6671 atm)*(3.8 L)/(0.082 L-atm/mol.K).(300.15 K) = 0.25739 mole ≈ 0.2574 mole.

The number of mole(s) of the ideal gas taken is 0.2574 mole (ans).