
A balloon contains $500{{m}^{3}}$of He, at ${{27}^{0}}C$ and 1 atmospheric pressure. The volume of He, at $-{{0.3}^{0}}C$and 0.5 atmospheric pressure will be
A. $700{{m}^{3}}$
B. $900{{m}^{3}}$
C. $1000{{m}^{3}}$
D. $500{{m}^{3}}$
Answer
584.7k+ views
Hint: The ideal gas equation is applicable for this question. Write down all the given values according to the specific symbol and then solve the question, it is a better way to solve thermodynamic problems. Use the ideal gas equation to find the value of volume.
Complete step-by-step answer:
Let us consider ${{V}_{1}},{{P}_{1}},{{T}_{1}}\text{ and }{{V}_{2}},{{P}_{2}},{{T}_{2}}$ be the volume, pressure and temperature of He at initial and final conditions respectively.
We know that the ideal gas equation, $PV=nRT$. Here n is the number of moles of the gas and R is universal gas constant.
For 1 mole of gas it becomes, $PV=RT$
$\dfrac{PV}{T}=R$
Now, we write the above equation for initial and final condition as,
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=R$and $\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}=R$
From the both equations we can conclude that,
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$
It is given that, ${{V}_{1}}=500{{m}^{3}}$ , ${{P}_{1}}=1atm$ , ${{T}_{1}}={{27}^{0}}C=300k$ , ${{P}_{2}}=0.5atm$ , ${{T}_{2}}=-{{3}^{0}}C=270k$ ,
Here, we have to find the final volume of He.
Putting all these values in above equation we get,
$\begin{align}
& {{V}_{2}}=\dfrac{270\times 1\times 500}{300\times 0.5} \\
& {{V}_{2}}=900{{m}^{3}} \\
\end{align}$
Hence the volume of He at $-{{3}^{0}}C$ and 0.5 atmospheric pressure will be $900{{m}^{3}}$.
Correct option is B.
Additional information:
There are basic gas laws in thermodynamic,
i. Charles’s law: $V\propto T$ , volume is directly proportional to temperature at constant pressure
ii. Boyle’s law: $V\propto \dfrac{1}{P}$ , volume is inversely proportional to pressure at constant temperature
iii. Avogadro’s law: $V\propto n$ , Volume is directly proportional to number of moles at constant pressure and temperature
Note: Write down the given values with specific symbols. Convert all the units in the same unit system if it is not given the same in question. Always don’t forget to convert the unit of temperature from degree Celsius to the Kelvin.
Complete step-by-step answer:
Let us consider ${{V}_{1}},{{P}_{1}},{{T}_{1}}\text{ and }{{V}_{2}},{{P}_{2}},{{T}_{2}}$ be the volume, pressure and temperature of He at initial and final conditions respectively.
We know that the ideal gas equation, $PV=nRT$. Here n is the number of moles of the gas and R is universal gas constant.
For 1 mole of gas it becomes, $PV=RT$
$\dfrac{PV}{T}=R$
Now, we write the above equation for initial and final condition as,
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=R$and $\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}=R$
From the both equations we can conclude that,
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$
It is given that, ${{V}_{1}}=500{{m}^{3}}$ , ${{P}_{1}}=1atm$ , ${{T}_{1}}={{27}^{0}}C=300k$ , ${{P}_{2}}=0.5atm$ , ${{T}_{2}}=-{{3}^{0}}C=270k$ ,
Here, we have to find the final volume of He.
Putting all these values in above equation we get,
$\begin{align}
& {{V}_{2}}=\dfrac{270\times 1\times 500}{300\times 0.5} \\
& {{V}_{2}}=900{{m}^{3}} \\
\end{align}$
Hence the volume of He at $-{{3}^{0}}C$ and 0.5 atmospheric pressure will be $900{{m}^{3}}$.
Correct option is B.
Additional information:
There are basic gas laws in thermodynamic,
i. Charles’s law: $V\propto T$ , volume is directly proportional to temperature at constant pressure
ii. Boyle’s law: $V\propto \dfrac{1}{P}$ , volume is inversely proportional to pressure at constant temperature
iii. Avogadro’s law: $V\propto n$ , Volume is directly proportional to number of moles at constant pressure and temperature
Note: Write down the given values with specific symbols. Convert all the units in the same unit system if it is not given the same in question. Always don’t forget to convert the unit of temperature from degree Celsius to the Kelvin.
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