Question

A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has a volume of 160.0 L and a pressure of 109.0 atm at 25.0 °C. After a while she notices that the valve has not been closed properly. The pressure had dropped to 99.0 atm. (The tank is still at 25.0 °C.) How many moles of gas has she lost?

Answer 1

Answer-

Given,

Volume of Helium Tank = 160.0 L

Initial Pressure (P1) of Tank = 109.0 atm

Final Pressure (P2) of Tank = 99.0 atm

Temperature of Tank = 25 ^$\degree$C or 298.15 K   [0°C + 273.15 = 273.15K]

moles lost due to leakage = ?

We know that,

PV = nRT

where,

P = Pressure

V = Volume

T = Temperature

n = Moles

R = Gas Constant (0.08206 L$\cdot$atm$\cdot$mol^-1$\cdot$K^-1)

So,

n = PV/RT

Initial moles (n1) = (P1*V)/(RT)

Final moles (n2) = (P2*V)/(RT)

Moles lost = Initial moles - Final Moles

Moles lost = (P1*V)/(RT) - (P2*V)/(RT)

Moles lost = (P1 - P2) * [V/(RT)]

Put the values,

Moles lost = (109.0 atm - 99.0 atm) * [160.0 L /(0.08206 L$\cdot$atm$\cdot$mol^-1$\cdot$K^-1 * 298.15 K)]

Moles lost = 65.4 mol [Answer]