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  • The pressure of a monoatomic gas increases linearly from <img alt="\mathrm{4 \times 10^5 \mathrm{~N} / \mathrm{m}^2\: to \: 8 \times 10^5 \mathrm{~N} / \mathrm{m}^2}" src="https://entrancecorner.on

The pressure of a monoatomic gas increases linearly from \mathrm{4 \times 10^5 \mathrm{~N} / \mathrm{m}^2\: to \: 8 \times 10^5 \mathrm{~N} / \mathrm{m}^2}when its volume increases from \mathrm{0.2 \mathrm{~m}^3 \: to \: 0.5 \mathrm{~m}^3.} Calculate increase in internal energy
 

Option: 1

\mathrm{\Delta \mathrm{U}=4.8 \times 10^5 \mathrm{~J}}
 


Option: 2

\mathrm{\Delta U=4.8 \times 10^4 \mathrm{~J}}
 


Option: 3

\mathrm{\Delta U=6.8 \times 10^5 \mathrm{~J}}

 


Option: 4

\mathrm{\Delta U=4.8 \times 10^6 \mathrm{~J}}


Answers (1)

best_answer

Change in internal energy of a gas is given by

\mathrm{ \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=\frac{\mathrm{nR} \Delta \mathrm{T}}{\gamma-1}=\frac{\mathrm{P}_{\mathrm{F}} \mathrm{V}_{\mathrm{F}}-\mathrm{P}_{\mathrm{I}} \mathrm{V}_1}{\gamma-1} }

As the gas is monoatomic, \gamma=5 / 3

\mathrm{\text { So } \Delta U=\frac{10^5(8 \times 0.5-4 \times 0.2)}{\left[\frac{5}{3}-1\right]}=\frac{3}{2} \times 10^5(4-0.8) }

\mathrm{ \Rightarrow \quad \Delta U=4.8 \times 10^5 \mathrm{~J}}



 

Posted by

Gautam harsolia

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