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A high altitude balloon contains 6.81 g of helium in {1.16}{10

Question

A high altitude balloon contains 6.81 g of helium in $1.16 \times 10^{4} L at - 23^{\circ} C .$ Assuming ideal gas behaviour, how many grams of helium would have to be added to increase the pressure to $4.0 \times 10^{- 3}$ atm?

1.27 g

1.58 g

2.68 g

2.23 g

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Michael Discussed

A high altitude balloon contains 6.81 g of helium in

1.16 \times 10^{4} L at - 23^{\circ} C .

Assuming ideal gas behaviour, how many grams of helium would have to be added to increase the pressure to

4.0 \times 10^{- 3}

atm?

8 mins ago

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8 mins ago

Text solutionVerified

Answer: dSolution: (d)

P_{1} = \frac{6.81}{4} \times \frac{0.0821 \times 250}{1.1 \times 104}

= 3.012 \times 10^{- 3}

P_{f} = 4 \times 10^{- 3}

△ P ↑= 1 \times 10^{- 3}

n_{He-added} = \frac{10 ^{- 3} \times 1.16 \times 10 ^{4}}{0.0821 \times 250}

w t = n_{He} \times 4 = 2.24 gm .

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Question Text	A high altitude balloon contains 6.81 g of helium in $1.16 \times 10^{4} L at - 23^{\circ} C .$ Assuming ideal gas behaviour, how many grams of helium would have to be added to increase the pressure to $4.0 \times 10^{- 3}$ atm?
Answer Type	Text solution:1
Upvotes	150