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A high altitude balloon contains $ 6.81g $ of helium in $ 1.16 \times {10^4}L $ at $ - 23^\circ C $ . Assuming ideal gas behavior, how many grams of helium would have to be added to increase the pressure to $ 4.0 \times {10^{ - 3}}atm $ ?
(a) $ 1.27 $
(b) $ 1.58 $
(c) $ 2.68 $
(d) $ 2.23 $

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Last updated date: 13th Jul 2024
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Answer
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Hint: To solve this question, we must consider an important law called the ideal gas law. From the ideal gas law, we will obtain the number of moles of the gas when pressure is $ 4.0 \times {10^{ - 3}}atm. $ From this, we can calculate the mass of the gas. To obtain the value of the number of grams of helium to be added, we must subtract $ 6.81g $ from the mass of gas we calculated earlier.

Complete answer:
The ideal gas equation is as follows:
 $ PV = nRT $
Where,
P is the pressure of the gas,
V is the volume occupied by the gas,
N is the number of moles or amount of substance,
R is the universal gas constant, $ R = 0.0821\,L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}} $ and
T is the temperature in Kelvin scale.
We are given:
 $ P = 4.0 \times {10^{ - 3}}atm $
 $ V = 1.16 \times {10^4}L $
 $ T = - 23^\circ = - 23 + 273K = 250K $
We know that: $ {\text{Number of moles (}}n{\text{) = }}\dfrac{{{\text{Mass of the substance}}}}{{{\text{Molar mass}}}} = \dfrac{w}{M} $
We are given: $ w = 6.81g $
M (for Helium) $ = \,4\,gmo{l^{ - 1}} $
From given values of P, V and T,
 $ PV = nRT $
 $ n = \dfrac{{PV}}{{RT}} $
 $ = \dfrac{{4.0 \times {{10}^{ - 3}}atm \times 1.16 \times {{10}^4}L}}{{0.0821\,L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}} \times 250K}} $
 $ = 2.26\,moles $
Since $ n = \dfrac{w}{M} $
 $ w = n \times M $
 $ = 2.26 \times 4 $
 $ = 9.04g $

Then, grams of helium to be added to increase the pressure to $ 4.0 \times {10^{ - 3}}atm $ will be
 $ = 9.04 - 6.81 $
 $ = 2.23g $
The correct answer is (d) $ 2.23. $

Note:
The ideal gas law gives the equation for a gas which is hypothetically ideal as it does not exist in real life. It can be used to give an approximation to the behaviour of other gases. It has several limitations. On considering both molecular size and intermolecular attractions, it is accurate for monatomic gases at high temperatures and low pressures. The van der Waals equation accounts for deviations from ideal behaviour caused by molecular size and intermolecular forces.