A monoatomic ideal gas goes through a process p=p0- α V where p0 and α are positive constants and V is its volume. At what volume will the entropy of gas be maximum?

Question

A monoatomic ideal gas goes through a process p=p0- α V where p0 and α are positive constants and V is its volume. At what volume will the entropy of gas be maximum? (A) (5 p0/6 α) (B) (p0/2 α) (C) (p0/4 α) (D) (5 p0/8 α)

Tardigrade Admin · Accepted Answer

d s=n Cv d T+P d V=0 n R (d T/d V)+(p0-α V)=0 pV=n R T p0 V-α V2=n R T p0-2 a V=n R (d T/d V) -(p0-α V)(&gamma;-1)=p0-2 a V -p0(&gamma;-1)+a(&gamma;-1) V=P0-2 a V p0 V=α V(&gamma;+1) V=(P0 &gamma;/a(&gamma;+1)) V=(P0 × (5/3)/a((5)3+1)) =(5P0/P0)

Q. A monoatomic ideal gas goes through a process $p = p_{0} -$ $α V$ where $p_{0}$ and $α$ are positive constants and $V$ is its volume. At what volume will the entropy of gas be maximum?

$\frac{5 p _{0}}{6 α}$

$\frac{p _{0}}{2 α}$

$\frac{p _{0}}{4 α}$

$\frac{5 p _{0}}{8 α}$

Q. A monoatomic ideal gas goes through a process p=p0​− αV where p0​ and α are positive constants and V is its volume. At what volume will the entropy of gas be maximum?

6α5p0​​

2αp0​​

4αp0​​

8α5p0​​

ds=nCv​dT+PdV=0 nRdVdT​+(p0​−αV)=0 pV​=nRT p0​V−αV2=nRT p0​−2aV=nRdVdT​ −(p0​−αV)(γ−1)=p0​−2aV −p0​(γ−1)+a(γ−1)V=P0​−2aV p0​V=αV(γ+1) V=a(γ+1)P0​γ​ V=a(35​+1)P0​×35​​=P0​5P0​​

Q. A monoatomic ideal gas goes through a process $p = p_{0} -$ $α V$ where $p_{0}$ and $α$ are positive constants and $V$ is its volume. At what volume will the entropy of gas be maximum?

$\frac{5 p _{0}}{6 α}$

$\frac{p _{0}}{2 α}$

$\frac{p _{0}}{4 α}$

$\frac{5 p _{0}}{8 α}$