Answer:
(a)
For
isothermal process, \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
For
adiabatic process, \[{{P}_{1}}V_{1}^{\gamma }\,=\,{{P}_{2}}V_{2}^{\gamma }\]
For
isothermal process,
For
adiabatic process, \[{{P}_{2a}}\,={{P}_{1}}\,{{\left(
\frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}\]\[={{2}^{\gamma
}}\,{{P}_{1}}\]
\[\therefore
\] \[\frac{{{P}_{2a}}}{{{P}_{2i}}}\,={{2}^{\gamma -1}}\]
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