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The density \rho vs pressure \mathrm{P} graph of an ideal monoatomic gas undergoes a process as shown in the figure. Find the total work done during the complete process. Molecular weight of gas is \mathrm{M} and number of moles is \mathrm{1 }


 

Option: 1

\mathrm{\frac{P_0 M}{\rho_0}(1-\ln 3)}
 


Option: 2

\mathrm{\frac{P_0 M}{\rho_0}(2-\ln 3)}
 


Option: 3

\mathrm{\frac{P_0 M}{\rho_0}(3-\ln 3)}
 


Option: 4

None of these


Answers (1)

best_answer

\mathrm{W_{12}=n R T_1 \ln \frac{1}{3}=n R \frac{P_0 M}{\rho_0 R}(-\ln 3) }

         \mathrm{ =-n \frac{P_0 M}{\rho_0} \ln 3=-\frac{P_0 M}{\rho_0} \ln 3 }

\mathrm{\mathrm{W}_{23} =3 \mathrm{P}_0\left(\mathrm{~V}_3-\mathrm{V}_2\right) }

\mathrm{ =3 \mathrm{P}_0\left(\frac{\mathrm{M}}{\rho_0}-\frac{\mathrm{M}}{3 \rho_0}\right)=\frac{2 \mathrm{P}_0 \mathrm{M}}{\rho_0}=\frac{2 \mathrm{P}_0 \mathrm{M}}{\rho_0}-\frac{\mathrm{P}_0 \mathrm{M}}{\rho_0} \ln 3=\frac{\mathrm{P}_0 \mathrm{M}}{\rho_0}(2-\ln 3) }



 

Posted by

Irshad Anwar

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