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  • The pressure of a monoatomic gas increases linearly from <img alt="4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?4%20%5Ctimes%2010%5E%7B5%7D%20%

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

Question :  increase in internal energy

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Change in internal energy of a gas is given by

\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}_{1} \mathrm{~V}_{1}}{\gamma-1}
As the gas is monoatomic, \gamma=5 / 3

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

Posted by

Ritika Harsh

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