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2 moles of an ideal monatomic gas is expanded according to relation \mathrm{PT = constant} from its initial state \mathrm{ (P_0V_0)} to the final state, due to which its pressure becomes half of the initial pressure. The change in internal energy is

Option: 1

\mathrm{\frac{3}{4} \mathrm{P}_0 \mathrm{~V}_0}


 


Option: 2

\mathrm{\frac{3}{2} P_0 V_0}
 


Option: 3

\mathrm{\frac{9}{2} \mathrm{P}_0 \mathrm{~V}_0}
 


Option: 4

\mathrm{\frac{5}{2} \mathrm{P}_0 \mathrm{~V}_0}


Answers (1)

best_answer

\mathrm{ \text { PT = constant } }

\mathrm{ \qquad P_0 T_i=\frac{P_0}{2} T_f }

\mathrm{ T_f=2 T_i }

Also, \mathrm{ P_0 V_0=2 R . T_i }

\mathrm{ T_i=\frac{P_0 V_0}{2 R} }          .......(1)

\mathrm{P_f V_f=n R . T_f }

\mathrm{ \frac{P_0}{2} V_f=[2][R]\left[2 T_f\right] }      .........(2)

\mathrm{\Rightarrow T_f=\frac{P_0 V_0}{R} }

\mathrm{ \therefore \Delta U=\frac{3}{2} n R \Delta T=\frac{3}{2} P_0 V_0}





 

Posted by

avinash.dongre

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