A gaseous mixture of 2 moles of A, 3 moles of B, 5 moles of C and 10 moles of D is contained in a vessel. Assuming that gases are ideal and the partial pressure of C is 1.5 atm, the total pressure is

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A gaseous mixture of 2 moles of A, 3 moles of B, 5 moles of C and 10 moles of D is contained in a vessel. Assuming that gases are ideal and the partial pressure of C is 1.5 atm, the total pressure is

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3 atm

15 at m

6 atm

9 atm

answer is B.

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Detailed Solution

Given data is :

$n_{A} = 2, n_{B} = 3, n_{c} = 5, n_{D} = 10, p_{c} = 1.5$ Mole fraction of C, $x_{C} = \frac{n_{C}}{n_{a} + n_{b} + n_{c} + n_{d}}$

$= \frac{5}{2 + 3 + 5 + 10}$

$x_{c} = \frac{5}{20}$

We have a formula that

$P_{C} = x_{c} P_{total}$

$1.5 = \frac{5}{20} \times P$

$P_{total} = 1.5 = \frac{5}{20} = 6$

Total pressure, $P_{total} = 6 atm$

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