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An ideal monoatomic gas at temperature 27^{\circ} \mathrm{C} and pressure 10^6 \mathrm{~N} / \mathrm{m}^2 occupies a volume of 10 litre. 10,000 of heat is added to the system without changing the volume. The final temperature of the gas is: Given : \mathrm{\mathrm{R}=8.31 \mathrm{~J} /(\mathrm{mol}-\mathrm{K}) \: and \: \mathrm{J}=4.18 \mathrm{~J} / \mathrm{Cal}}
 

Option: 1

780^{\circ} \mathrm{C}


 


Option: 2

810^{\circ} \mathrm{C}
 


Option: 3

840^{\circ} \mathrm{C}
 


Option: 4

860^{\circ} \mathrm{C}


Answers (1)

best_answer

For \mu mole of gas, we have

\mathrm{PV}=\mu \mathrm{RT}

\text { Here } \mathrm{p}=10^6 \mathrm{nt} / \mathrm{m}^2, \mathrm{~V}=10 \text { litre }=10^{-2} \mathrm{~m}^3 \text { and } \mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}

\mathrm{ \therefore \quad \mu=\frac{\mathrm{pV}}{\mathrm{RT}}=\frac{10^6 \times 10^{-2}}{8.31 \times 300}=4.0}

For a "monoatomic" gas, \mathrm{\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}}(by kinetic theory and equipartition of energy).

Thus,\mathrm{C_V =\frac{3}{2} \times 8.31 \mathrm{~J} / \mathrm{mol}-\mathrm{K} }

\mathrm{ =\frac{3}{2} \times \frac{8.31}{4.18} \approx 3 \mathrm{Cal} /(\text { mole }-\mathrm{K}) .}

Let \mathrm{\Delta \mathrm{T}} be the rise in temperature when \mathrm{\mu} mole of the gas is given \mathrm{\mathrm{Q}} cal of heat at constant volume. Then

\mathrm{ \mathrm{Q}=\mu \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T} }

\mathrm{ \text { or } \quad \Delta \mathrm{T}=\frac{\mathrm{Q}}{\mu \mathrm{C}_{\mathrm{v}}}=\frac{10,000 \mathrm{cal}}{4.0 \mathrm{~mole} \times 3 \mathrm{cal}(\mathrm{mole}-\mathrm{K})}=833 \mathrm{~K} }

\mathrm{ \therefore \quad \text { final temperature of the gas is } }

\mathrm{ \mathrm{T}+\Delta \mathrm{T}=300+833=1133 \mathrm{~K}=860^{\circ} \mathrm{C} }



 

Posted by

Divya Prakash Singh

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