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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]
[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]
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