A gas cylinder contains 2.0 mol of gas X and 6.0 mol of gas Y at a total pressure of 2.1 atm. What is the partial pressure of gas Y?

Use the formula: [tex]\(\frac{P_a}{P_T} = \frac{n_a}{n_T}\)[/tex]

A. 0.50 atm
B. 1.6 atm
C. 2.1 atm
D. 2.8 atm



Answer :

To find the partial pressure of gas [tex]\( Y \)[/tex], we can use Dalton's Law of Partial Pressures. This law states that the partial pressure of each gas in a mixture is proportional to its mole fraction in the mixture. The formula is given as follows:

[tex]\[ \frac{P_a}{P_T} = \frac{n_a}{n_T} \][/tex]

Where:
- [tex]\( P_a \)[/tex] is the partial pressure of gas [tex]\( a \)[/tex]
- [tex]\( P_T \)[/tex] is the total pressure
- [tex]\( n_a \)[/tex] is the number of moles of gas [tex]\( a \)[/tex]
- [tex]\( n_T \)[/tex] is the total number of moles of gases

Given:
- [tex]\( n_X = 2.0 \)[/tex] moles (gas [tex]\( X \)[/tex])
- [tex]\( n_Y = 6.0 \)[/tex] moles (gas [tex]\( Y \)[/tex])
- [tex]\( P_T = 2.1 \)[/tex] atm (total pressure)

First, let's compute the total number of moles [tex]\( n_T \)[/tex]:

[tex]\[ n_T = n_X + n_Y = 2.0 + 6.0 = 8.0 \][/tex]

Next, we use the mole fraction of gas [tex]\( Y \)[/tex] to determine its partial pressure:

[tex]\[ \frac{P_Y}{P_T} = \frac{n_Y}{n_T} \][/tex]

Solving for [tex]\( P_Y \)[/tex]:

[tex]\[ P_Y = \left(\frac{n_Y}{n_T}\right) \cdot P_T \][/tex]

Substituting the given values:

[tex]\[ P_Y = \left(\frac{6.0}{8.0}\right) \cdot 2.1 \][/tex]

[tex]\[ P_Y = 0.75 \cdot 2.1 \][/tex]

[tex]\[ P_Y = 1.575 \, \text{atm} \][/tex]

Thus, the partial pressure of gas [tex]\( Y \)[/tex] is approximately [tex]\( 1.575 \, \text{atm} \)[/tex].

From the given choices, [tex]\( 1.575 \, \text{atm} \)[/tex] is closest to [tex]\( 1.6 \, \text{atm} \)[/tex].

Therefore, the answer is:
[tex]\[ 1.6 \, \text{atm} \][/tex]