Question

Container A holds 717 mL of an ideal gas at 2.80 atm. Container B holds 174 mL of a different ideal gas at 4.30 atm. Container A and container B are glass spheres connnected by a tube with a stopcock. Container A is larger than container B. If the gases are allowed to mix together, what is the resulting pressure

Answer 1

Answer:

3.09 atm

Explanation:

Given that:

Volume of container A = 717 mL

Pressure of container A = 2.80 atm

Volume of container B = 174 mL

Pressure of container B = 4.30 atm

Now, if the valve are being removed and the gases are allowed to mix together; then

The total final pressure can be calculated by using the formula:

$P_f = \dfrac{P_AVA+P_BV_B}{V_A+V_B}$

$P_f = \dfrac{2.80*717+4.30*174}{717+174}$

$P_f = \dfrac{2007.6+748.2}{891}$

$P_f = \dfrac{2755.8}{891}$

$\mathbf{P_f =3.09 \ atm}$

Answer 2

The absence of the intermolecular attraction between the atoms or molecules is called an ideal gas.  The pressure of the resultant mixture is 3.09 atm.

What is the relation between pressure and volume?

The relationship between the pressure and the volume of the gas is given by Boyle's law which states the inverse proportionality relation with the volume.

Given,

The volume of container A = 717 mL

The volume of container B = 174 mL

The pressure of container A = 2.80 atm

The pressure of container B = 4.30 atm

The total final pressure of the ideal gas of the mixture is calculated by:

$\begin{aligned} \rm P_{f} &= \rm \dfrac{P_{A}V_{A}+ P_{B}V_{B}}{V_{A}+V_{B}}\\\\&= \dfrac{2007.6 +748.2}{891}\\\\&= 3.09 \;\rm atm\end{aligned}$

Therefore, the total pressure of the mixture is 3.09 atm.

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