Question

a monoatomic gas at a pressure P, having volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V . The final pressure of the gas is

Answer 1

$$\begin{gathered} Gaseqnforisothermalprocessis, \\ {P}_1{V}_1=P_2{V}_2 \\ or{P}_2=\frac{P_1{V}_1}{V_2},puttingthevaluesofthesedata,weget, \\ {P}_2=\frac{PV}{2V}=\frac{P}{2}. \\ Nowthegasisexpandedadiabatically,soeqnofadiabaticprocessis, \\ {P}_2{V_2}^{\gamma }=P_0{V_0}^{\gamma },againputtingthevaluesofthegivendata,weget, \\ \frac{P}{2}\times {\left(2v\right)}^{\gamma }=p_0\times {\left(16v\right)}^{\gamma } \\ {P}_0=\frac{P}{2}\times {\left(\frac{2}{16}\right)}^{\gamma }=\frac{P}{2}\times {\left(\frac{1}{8}\right)}^{\gamma }. \\ Butformonoatomicgas\gamma =\frac{5}{3},so \\ {P}_0=\frac{P}{2}\times {\left(\frac{1}{2}\right)}^{3\times \frac{5}{3}}=\frac{P}{2}\times {\left(\frac{1}{2}\right)}^5=\frac{P}{2}\times \frac{1}{32}=\frac{P}{64}. \\ Thereforefinalpressure{P}_0=\frac{P}{64}. \end{gathered}$$