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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: work done by the gas

Option: 1

2.8 \times 10^{5} \mathrm{~J}


Option: 2

1.8 \times 10^{6} \mathrm{~J}


Option: 3

1.8 \times 10^{5} \mathrm{~J}


Option: 4

1.8 \times 10^{2} \mathrm{~J}


Answers (1)

best_answer

Work done by the gas,


\Delta \mathrm{W}=\int \mathrm{PdV}=\text { area under } \mathrm{P}-\mathrm{V} \text { curve }
=\mathrm{P}_{1}\left(\mathrm{~V}_{\mathrm{F}}-\mathrm{V}_{\mathrm{l}}\right)+\frac{1}{2}\left(\mathrm{P}_{\mathrm{F}}-\mathrm{P}_{\mathrm{I}}\right) \times\left(\mathrm{V}_{\mathrm{F}}-\mathrm{V}_{\mathrm{l}}\right)
=\frac{1}{2}\left(\mathrm{~V}_{\mathrm{F}}-\mathrm{V}_{\mathrm{l}}\right)\left(\mathrm{P}_{\mathrm{F}}+\mathrm{P}_{\mathrm{I}}\right)
=\frac{1}{2}(0.5-0)(8+4) \times 10^{5}=1.8 \times 10^{5} \mathrm{~J}

Posted by

Ajit Kumar Dubey

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