question_answer 29)

                  Consider a cycle tyre being filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump \[\Delta V(=V)\]of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from \[{{P}_{1}}\] to \[{{P}_{2}}\]?                

Answer:

                  In case of adiabatic process, \[P{{V}^{\gamma }}=\] constant                 Now \[P{{(V+\Delta V)}^{\gamma }}=(P+\Delta P){{V}^{\gamma }}\]                 \[=\,P\left( 1+\frac{\Delta P}{P} \right)\,{{V}^{\gamma }}\]                 or \[\frac{{{(V+\Delta V)}^{\gamma }}}{{{V}^{\gamma }}}\,=\,\left( 1+\,\frac{\Delta P}{P} \right)\]                 or \[{{\left( \frac{V+\Delta V}{V} \right)}^{\gamma }}\,=\,1+\,\frac{\Delta P}{P}\]                 or \[{{\left( 1+\,\frac{\Delta V}{V} \right)}^{\gamma }}\,=\,1+\,\frac{\Delta P}{P}\]                 Since \[\Delta V<<V,\] so using Binomial theorem, we get                 \[1+\,\gamma \,\frac{\Delta V}{V}\,\,=1+\,\frac{\Delta P}{P}\]                 or \[\gamma \,\frac{\Delta V}{V}=\,\frac{\Delta V}{P}\] or \[\frac{\Delta V}{\Delta P}=\,\frac{V}{\gamma P}\]                 or \[\frac{dV}{dP}\,=\,\frac{V}{\gamma P}\] or \[dV=\,\frac{Vdp}{\gamma P}\]                 Work done during adiabatic process,             \[W=\,\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{P\,dV=\,\,}\,\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\frac{PV\,dP}{\gamma P}\,}\]                 \[=\frac{V}{\gamma }\,\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{dP}\] \[(\because \,V=\,fixed)\]                 \[=\,\frac{V\,({{P}_{2}}-{{P}_{2}})}{\gamma }\].