A) \[mg\left( 1+\frac{h}{d} \right)\]
B) \[mg{{\left( 1+\frac{h}{d} \right)}^{2}}\]
C) \[mg\left( 1-\frac{h}{d} \right)\]
D) \[mg\]
Correct Answer: B
Solution :
When 1 mole of ideal gas is expanded adiabatically from an initial volume \[W=\frac{1}{\gamma -1}\left[ \frac{{{P}_{i}}{{V}_{i}}}{V_{i}^{\gamma -1}}-\frac{{{p}_{f}}{{V}_{f}}}{V_{f}^{\gamma -1}} \right]\] to final volume \[=\frac{1}{\gamma -1}\left[ {{P}_{i}}{{V}_{i}}-{{P}_{f}}{{V}_{f}} \right]\], then the external work done by the gas is \[{{T}_{i}}\] Where \[{{T}_{f}}\] is instantaneous pressure of the gas during infinitesimal expansion\[{{P}_{i}}{{V}_{i}}=R\,\,{{T}_{i}}\,\,and\,\,{{P}_{f}}{{V}_{f}}=R\,\,T\]. In adiabatic expansion \[\therefore \] \[W=\frac{R}{\gamma -1}\left( {{T}_{i}}-{{T}_{f}} \right)\] \[{{C}_{P}}-{{C}_{V}}=R\] \[\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}\] \[\therefore \] Where, \[W=\frac{{{C}_{p}}-{{C}_{V}}}{\frac{{{C}_{P}}-{{C}_{V}}}{{{C}_{V}}}}\left( {{T}_{i}}-{{T}_{f}} \right)={{C}_{V}}\left( {{T}_{i}}-{{T}_{f}} \right)\] \[{{T}_{i}}\] \[{{T}_{1}}\] If \[{{T}_{f}}\] and \[{{T}_{2}}\] be absolute temperatures of the gas, then \[\therefore \] \[W={{C}_{V}}\left( {{T}_{1}}-{{T}_{2}} \right)\] \[F\left( x \right)=-\frac{dU}{dx}\] From Mayor?s formula \[U=A-B{{x}^{2}}\] and \[\therefore \] \[\frac{dU}{dx}=-2Bx\] \[\therefore \] Given, \[F\left( x \right)=-2Bx\]=\[\Rightarrow \] and \[F\left( x \right)\propto x\]=\[x\] \[\therefore \]\[mg\left( h+d \right)=F\times d\] Note: Work done depends only one the initial and final temperature of the gas
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