A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume of the balloon if 0.20 moles of gas are added?

V = 1.6 L

Explanation:

assuming ideal gas:

PV = RTn

∴ R = 0.082 atm. L/K. mol

∴ V1 = 1.50 L

∴ n1 = 3.00 mol

∴ T1 = 25°C ≅ 298 K

⇒ P1 = (RT1n1) / (V1) = ((0.082 atm. L/K. mol) (298 K) (3.00 mol)) / (1.50 L)

⇒ P1 = 48.872 atm

with pressure and temperature held constant:

∴ T2 = T1 = 298 K

∴ P2 = P1 = 48.872 atm

∴ n2 = 0.20 mol + 3.00 mol = 3.20 mol

⇒ V2 = (RT2n2) / P2

⇒ V2 = ((0.082 atm. L/K. mol) (298 K) (3.20 mol)) / (48.872 atm)

⇒ V2 = 1.6 L

Answer: 1.6L

Explanation:

V1 = 1.50 L,

V2 = ?

n1 = 3mol

n2 = 3 + 0.2 = 3.2mol

From PV = nRT

V1 / n1 = V2/n2

1.5/3 = V2 / 3.2

V2 = (1.5/3/) x 3.2 = 1.6L