For a reversible thermodynamic process for monoatomic gas PVx= constant. If for this process x=(CP/CV), then heat capacity for the process is :-

Question

For a reversible thermodynamic process for monoatomic gas PVx= constant. If for this process x=(CP/CV), then heat capacity for the process is :- (A) (3 R/2) (B) (5 R/2) (C) 0 (D) ∞

Tardigrade Admin · Accepted Answer

x=(CP/CV)=&gamma; so, process is an adiabatic process. So, heat capacity =0

Q. For a reversible thermodynamic process for monoatomic gas $P V^{x} =$ constant. If for this process $x = \frac{C _{P}}{C _{V}},$ then heat capacity for the process is :-

$\frac{3 R}{2}$

$\frac{5 R}{2}$

$0$

$\infty$

$x = \frac{C _{P}}{C _{V}} = γ$ so, process is an adiabatic process.
So, heat capacity $= 0$

Q. For a reversible thermodynamic process for monoatomic gas PVx= constant. If for this process x=CV​CP​​, then heat capacity for the process is :-

23R​

25R​

0

∞

x=CV​CP​​=γ so, process is an adiabatic process. So, heat capacity =0

Q. For a reversible thermodynamic process for monoatomic gas $P V^{x} =$ constant. If for this process $x = \frac{C _{P}}{C _{V}},$ then heat capacity for the process is :-

$\frac{3 R}{2}$

$\frac{5 R}{2}$

$0$

$\infty$

$x = \frac{C _{P}}{C _{V}} = γ$ so, process is an adiabatic process.
So, heat capacity $= 0$