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Q.

A high altitude balloon contains 6.81 g of helium in 1.16 x 10⁴ L at - 23^oC. Assuming ideal gas behavior, how many grams of helium would have to be added to increase the pressure to 4.0 x 10^-3atm ?

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a

1.27

b

1.58

c

2.68

d

2.23

answer is D.

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Detailed Solution

$P_{i} = \frac{nRT}{V} = \frac{6 .81}{4} \times \frac{0 .0821 \times 250}{1 .16 \times 10^{4}} \begin{matrix} = 3 .012 \times 10^{- 3} \\ P_{f} = 4 \times 10^{- 3} \\ ΔP ↑= 1 \times 10^{- 3} \end{matrix} \begin{aligned} n_{He-added} = \frac{10^{- 3} \times 1 .16 \times 10^{4}}{0 .0821 \times 250} \\ wt = n_{He} \times 4 = 2 .24 gm \end{aligned}$

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