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  • During an experiment an ideal gas is found to obey an additional law <img alt="\mathrm{ \mathrm{VP}^2= }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cmathrm%7BVP%7D%5E2%3

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

Option: 1

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

 


Option: 2

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


Option: 3

\mathrm{T^{\prime}=\sqrt{5} T}
 


Option: 4

\mathrm{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 \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}_1}{\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

Rishi

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