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During an experiment an ideal gas is found to obey an additional law \mathrm{VP}^{2}= constant. The gas is initially at temp \mathrm{T} and volume \mathrm{V}. What will be temperature of the gas when it expands to a volume \mathrm{2 \mathrm{~V}}?

Option: 1

\mathrm{\quad T^{\prime}=\sqrt{4} T}


Option: 2

\mathrm{\quad T^{\prime}=\sqrt{2} T}


Option: 3

\mathrm{T^{\prime}=\sqrt{5} T}


Option: 4

\mathrm{\quad T^{\prime}=\sqrt{6} T}


Answers (1)

best_answer

According to the given problems

\mathrm{VP}^{2}=\text { constant }

From the gas law

\mathrm{\mathrm{PV}=\mathrm{nRT} }
\mathrm{\Rightarrow \quad \left(\frac{\mathrm{k}}{\sqrt{\mathrm{V}}}\right) \mathrm{V}=\mathrm{nRT} }
\mathrm{\Rightarrow \quad \sqrt{\mathrm{V}}=\left(\frac{\mathrm{nR}}{\mathrm{K}}\right) \mathrm{T} }
\mathrm{\therefore \quad \sqrt{\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}}=\left(\frac{\mathrm{T}}{\mathrm{T}_{2}}\right), \text { i.e, } \sqrt{\frac{\mathrm{V}}{2 \mathrm{~V}}}=\frac{\mathrm{T}}{\mathrm{T}^{\prime}}}
\mathrm{\Rightarrow \quad \mathrm{T}^{\prime}=\sqrt{2} \mathrm{~T}}

Posted by

Divya Prakash Singh

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