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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 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,

\mathrm{ \Delta W=\int P d V=\text { area under } P \text {-v curve }}

\mathrm{ =P_1\left(V_F-V_1\right)+\frac{1}{2}\left(P_F-P_1\right) \times\left(V_F-V_1\right) }

\mathrm{ =\frac{1}{2}\left(V_F-V_1\right)\left(P_F+P_1\right) }

\mathrm{ =\frac{1}{2}(0.5-0 .)(8+4) \times 10^5=1.8 \times 10^5 \mathrm{~J}}





 

Posted by

Sanket Gandhi

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