Question

A closed container holds 1.33 moles of gas A 2.14 moles of gas B 1.12 moles of gas C If the total pressure within the container is 5.28 atm, what is the partial pressure of gas B?

Answer 1

Answer:

2.48 atm

Explanation:

The partial pressure of the gas is equal to the product of the mole fraction and total pressure.

So,  

$P_{gas}=X_{gas}\times P$

Given that:-

Moles of A = 1.33 moles

Moles of B = 2.14 moles

Moles of C = 1.12 moles

Mole fraction of B can be calculated as:-

$X_B=\frac{n_B}{n_A+n_B+n_C}=\frac{2.14}{1.33+2.14+1.12}=0.47$

Total pressure = 5.28 atm

Thus, partial pressure of gas B = 0.47 x 5.28 atm = 2.48 atm

Answer 2

Answer:

The partial pressure of gas B is 2.46 atm

Explanation:

Step 1: Data given

moles of gas A = 1.33 moles

moles of gas B = 2.14 moles

moles of gas C = 1.12 mol

Total pressure = 5.28 atm

Step 2: Calculate the total number of moles

Total moles = moles gas A + moles gas B + moles gas C

Total moles = 1.33 + 2.14 + 1.12

Total moles = 4.59

Step 3: Calculate mol fraction

mol fraction gas A = 1.33 / 4.59 = 0.290

mol fraction gas B = 2.14 / 4.59 = 0.466

mol fraction gas C = 1.12 / 4.59 = 0.244

Step 4: Calculate partial pressures

Partial pressure gas A = 0.290 * 5.28 atm = 1.53 atm

Partial pressure gas B = 0.466 * 5.28 atm = 2.46 atm

Partial pressure gas C = 0.244 * 5.28 atm = 1.29 atm

The partial pressure of gas B is 2.46 atm