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Crank
3 years ago
5

Container A holds 722 mL 722 mL of an ideal gas at 2.40 atm. 2.40 atm. Container B holds 169 mL 169 mL of a different ideal gas

at 4.60 atm. 4.60 atm. If the gases are allowed to mix together, what is the resulting pressure?
Chemistry
1 answer:
iren [92.7K]3 years ago
6 0

Answer:

The resulting pressure is 2.81 atm

Explanation:

According to Dalton's Law of Partial Pressure, each of the gases (A and B) will exert their pressure independently. If we use Boyle's Law to calculate the pressure of each of the gases separately we have:

Pressure of gas A:

p1V1 = p2V2

p1 = 2.4 atm

V1 = 722 mL

V2 = 722 + 169 = 891 mL

p2 =?

Clearing p2:

p2 = (p1V1)/V2 = (2.4*722)/891 = 1.94 atm

Pressure of gas B:

p1 = 4.6 atm

V1 = 169 mL

V2 = 169+722 = 891 mL

p2=?

Clearing p:

p2 = (4.6*169)/891 = 0.87 atm

Dalton's expression for total partial pressures is equal to:

ptotal = pA + pB = 1.94+0.87 = 2.81 atm

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If 20.0g of CO2 and 4.4g of CO2
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The given question is incorrect. The correct question is as follows.

If 20.0 g of O_{2} and 4.4 g of CO_{2}  are placed in a 5.00 L container at 21^{o}C, what is  the pressure of this mixture of gases?

Explanation:

As we know that number of moles equal to the mass of substance divided by its molar mass.

Mathematically,   No. of moles = \frac{\text{mass}}{\text{molar mass}}

Hence, we will calculate the moles of oxygen as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

     Moles of O_{2} = \frac{20.0 g}{32 g/mol}

                            = 0.625 moles

Now,   moles of CO_{2} = \frac{4.4 g}{44 g/mol}

                                      = 0.1 moles

Therefore, total number of moles present are as follows.

Total moles = moles of O_{2} + moles of CO_{2}

                    = 0.625 + 0.1

                    = 0.725 moles

And, total temperature  will be:

                    T = (21 + 273) K = 294 K

According to ideal gas equation,  

                         PV = nRT

Now, putting the given values into the above formula as follows.

                P = \frac{nRT}{V}

                   = \frac{0.725 mol \times 0.08206 Latm/mol K \times 294 K}{5.00 L}

                    = \frac{17.491089}{5} atm

                    = 3.498 atm

or,                = 3.50 atm (approx)

Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.

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