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Question: A high altitude balloon contains \( 6.81g \) of helium in \( 1.16 \times {10^4}L \) at \( - 23^\circ...

A high altitude balloon contains 6.81g6.81g of helium in 1.16×104L1.16 \times {10^4}L at 23C- 23^\circ C . Assuming ideal gas behavior, how many grams of helium would have to be added to increase the pressure to 4.0×103atm4.0 \times {10^{ - 3}}atm ?
(a) 1.271.27
(b) 1.581.58
(c) 2.682.68
(d) 2.232.23

Explanation

Solution

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 4.0×103atm.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 6.81g6.81g from the mass of gas we calculated earlier.

Complete answer:
The ideal gas equation is as follows:
PV=nRTPV = 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=0.0821LatmK1mol1R = 0.0821\,L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}} and
T is the temperature in Kelvin scale.
We are given:
P=4.0×103atmP = 4.0 \times {10^{ - 3}}atm
V=1.16×104LV = 1.16 \times {10^4}L
T=23=23+273K=250KT = - 23^\circ = - 23 + 273K = 250K
We know that: Number of moles (n) = Mass of the substanceMolar mass=wM{\text{Number of moles (}}n{\text{) = }}\dfrac{{{\text{Mass of the substance}}}}{{{\text{Molar mass}}}} = \dfrac{w}{M}
We are given: w=6.81gw = 6.81g
M (for Helium) =4gmol1= \,4\,gmo{l^{ - 1}}
From given values of P, V and T,
PV=nRTPV = nRT
n=PVRTn = \dfrac{{PV}}{{RT}}
=4.0×103atm×1.16×104L0.0821LatmK1mol1×250K= \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}}
=2.26moles= 2.26\,moles
Since n=wMn = \dfrac{w}{M}
w=n×Mw = n \times M
=2.26×4= 2.26 \times 4
=9.04g= 9.04g

Then, grams of helium to be added to increase the pressure to 4.0×103atm4.0 \times {10^{ - 3}}atm will be
=9.046.81= 9.04 - 6.81
=2.23g= 2.23g
The correct answer is (d) 2.23.2.23.

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.