Question

You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar at 25°C and 2.75 atm pressure. A stopcock connecting the containers is opened and the gases are allowed to equilibrate between the two containers. What is the final pressure in the two containers if the temperature remains at 25°C? Assume ideal behavior.

Answer 1

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Thus, the expression for final pressure in the two containers will be:

$PV=P_1V_1+P_2V_2$

$P=\frac{P_1V_1+P_2V_2}{V}$

where,

$P_1$ = pressure of N₂ gas = 4.45 atm

$P_2$ = pressure of Ar gas = 2.75 atm

$V_1$ = volume of N₂ gas = 3.00 L

$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

$P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}$

$P=2.62atm$

Thus, the final pressure in the two containers is, 2.62 atm

Answer 2

The final pressure of the gases in the two containers is 3.77 atm.

The given parameters;

• initial volume of gas in the container, V₁ = 3 liters
• initial pressure, P₁ = 4.45 atm
• final volume of the gas, V₂ = 2 Liters
• final pressure, P₂ = 2.75 atm

Apply Boyles law to determine the final pressure of the equilibrium mixture at the given constant temperature;

$P_tV_t = P_1V_1 + P_2V_2\\\\P_t = \frac{P_1V_1 + P_2V_2}{V_t} \\\\P_t = \frac{(4,45\times 3) \ + (2.75\times 2)}{(2+ 3)} \\\\P_t = 3.77 \ atm$

Thus, the final pressure of the gases in the two containers is 3.77 atm.

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