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Question: A balloon contains \(500{{m}^{3}}\) of helium at \(27{}^\circ C\) and 1 atmosphere pressure. The vol...

A balloon contains 500m3500{{m}^{3}} of helium at 27C27{}^\circ C and 1 atmosphere pressure. The volume of helium at 3C-3{}^\circ C temperature and 0.5 atmosphere pressure will be
A. 500m3500{{m}^{3}}
B. 700m3700{{m}^{3}}
C. 900m3900{{m}^{{{3}^{{}}}}}
D. 1000m31000{{m}^{3}}

Explanation

Solution

As a very first step, one could note down all the given values from the question. You could then recall the ideal gas equation. Assume that the number of moles of the given helium gas remains constant and then you will get a proportionality relation for pressure, volume and temperature and hence find the answer.
Formula used:
Ideal Gas equation,
PV=nRTPV=nRT

Complete step by step answer:
In the question we are given a balloon containing helium. The initial conditions are given as,
P1=1atm{{P}_{1}}=1atm
V1=500m3{{V}_{1}}=500{{m}^{3}}
T1=27C=300K{{T}_{1}}=27{}^\circ C=300K
We are given another set of conditions where,
P2=0.5atm{{P}_{2}}=0.5atm
T2=3C=270K{{T}_{2}}=-3{}^\circ C=270K
We are supposed to find the volume of helium contained now.
Let recall the ideal gas equation given by,
PV=nRTPV=nRT
We know that R is the universal gas constant and then for a given number of moles of a gas, we could say that,
PVαTPV\alpha T
P1V1T1=P2V2T2\Rightarrow \dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}
Substituting the given values we see that,
1×500300=0.5×V2270\Rightarrow \dfrac{1\times 500}{300}=\dfrac{0.5\times {{V}_{2}}}{270}
V2=900m3\therefore {{V}_{2}}=900{{m}^{3}}
Therefore, we found the volume of helium under the given second set of conditions to be,
V2=900m3{{V}_{2}}=900{{m}^{3}}

So, the correct answer is “Option C”.

Note: A very important step that should be followed in questions related to this topic is that the temperature when given in degree Celsius should be converted to Kelvin scale by adding 273 to the given value in Celsius. Since, all the other quantities are given in SI units one shouldn’t worry about them.