Question

In an adiabatic process, pressure is increased by 2/3%. If nu = 3/2 then the volume in the process is decreased by about.

Answer 1

Step 1: Relation between $\mathrm{P},\mathrm{V}\mathrm{and}\mathrm{\gamma }$

In adiabatic process there is no transfer of heat or matter between the surroundings and environment.

${\mathrm{pV}}^{\mathrm{\gamma }}=\mathrm{constant}$

Here,

$\mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{v}}}=\frac{3}{2}$

Therefore,

${\mathrm{pV}}^{\frac{3}{2}}=\mathrm{constant}$

Step 2:Applying log to both the sides

$\mathrm{logp}+\frac{3}{2}\left(\mathrm{logV}\right)=\mathrm{logC}$

$\frac{∆\mathrm{\pi }}{\mathrm{p}}+\frac{3}{2}\left(\frac{∆\mathrm{V}}{\mathrm{V}}\right)=0$

Step 3: Decreased volume percentage

Therefore,

$\frac{∆\mathrm{V}}{\mathrm{V}}=-\frac{2}{3}\left(\frac{∆\mathrm{\pi }}{\mathrm{p}}\right)$

$\frac{∆\mathrm{V}}{\mathrm{V}}\mathrm{x}100$ $=-\left(\frac{2}{3}\right)\left(\right[\frac{∆\mathrm{\pi }}{\mathrm{p}}]\mathrm{x}100)$

$=-\left(\frac{2}{3}\right)\mathrm{x}\frac{2}{3}\%=-\frac{4}{9}\%$

Therefore, volume in this process decreases by $\frac{4}{9}\%$