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One mole of an ideal gas undergoes a process T=300+2V where T is temperature and V is volume. If volume of gas increases from 2\; m^{3} to 4\; m^{3} then work done by the gas in this process will be

 

Option: 1

300R \; In2


Option: 2

2+2 \; In2


Option: 3

300R \; In2-4R

 


Option: 4

300R \; In2+4R


Answers (1)

Answer (4)

T=300+2V

Work done W=\int pdv

          =\int \frac{\left ( 300+2V \right )}{V}Rdv \begin{bmatrix} PV=nRT\\ P=\frac{RT}{V} \end{bmatrix}

         =300 \; R\; In\left ( 4-2 \right )+2\left ( 4-2 \right )R

         =300 \; R\; In 2+4R

Posted by

Ramraj Saini

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