Question

Two glass bulbs, both at 298 K, are connected by a (closed) valve: The left bulb has a volume of 3.0 L and is filled with 145 mmHg of helium gas The right bulb has a volume of 2.0 L and is filled with 355 mmHg of argon gas When the valve is opened and the gases are allowed to mix, what will be the total pressure in the system?

Answer 1

First calculate mole of gas

mole of gas in left bulb

Ideal gas equation

PV = nRT             where, P = atm pressure= 145 mmHg = 0.19 atm,

V = volume in Liter = 3 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 298 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.19 X 3)/(0.08205X 298) = 0.0233 mole

now calulate mole in right bulb

Ideal gas equation

PV = nRT             where, P = atm pressure= 355 mmHg = 0.467 atm,

V = volume in Liter = 2 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 298 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.467 X 2)/(0.08205X 298) = 0.0382 mole

After mixing of both gas

total mole = 0.0233 + 0.0382 = 0.0615 mole

total volume = 5 L

Ideal gas equation

PV = nRT             where, P = atm pressure= ?

V = volume in Liter = 5 L

n = number of mole = 0.0615 mol

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 298 K

We can write ideal gas equation

P = nRT/V

Substitute value in above equation

P = 0.0615X 0.08205 X 298 / 5 = 0.3 atm

0.3 atm = 228 mmHg

total pressure in system = 228 mm Hg