The ratio of the adiabatic bulk modulus to the isothermal bulk modulus of a perfect gas is equal to (symbols have their usual meanings)

Question

The ratio of the adiabatic bulk modulus to the isothermal bulk modulus of a perfect gas is equal to (symbols have their usual meanings) (A)  Cp-CV  (B)  (CP/CV)  (C)  (CV/CP)  (D)  √(CP/CV)

Tardigrade Admin · Accepted Answer

Adiabatic bulk modulus =Badia=&gamma; P Isothermal bulk modulus =Biso=P ⇒ (Badia/Biso)=&gamma; =(CP/CV)

Q. The ratio of the adiabatic bulk modulus to the isothermal bulk modulus of a perfect gas is equal to (symbols have their usual meanings)

$C_{p} - C_{V}$

$\frac{C _{P}}{C _{V}}$

$\frac{C _{V}}{C _{P}}$

$\frac{C _{P}}{C _{V}}$

Adiabatic bulk modulus $= B_{a d ia} = γ P$ Isothermal bulk modulus $= B_{i so} = P$ $\Rightarrow$ $\frac{B _{a d ia}}{B _{i so}} = γ = \frac{C _{P}}{C _{V}}$

Q. The ratio of the adiabatic bulk modulus to the isothermal bulk modulus of a perfect gas is equal to (symbols have their usual meanings)

Cp​−CV​

CV​CP​​

CP​CV​​

CV​CP​​​

Adiabatic bulk modulus =Badia​=γP Isothermal bulk modulus =Biso​=P ⇒ Biso​Badia​​=γ=CV​CP​​

$C_{p} - C_{V}$

$\frac{C _{P}}{C _{V}}$

$\frac{C _{V}}{C _{P}}$

$\frac{C _{P}}{C _{V}}$

Adiabatic bulk modulus $= B_{a d ia} = γ P$ Isothermal bulk modulus $= B_{i so} = P$ $\Rightarrow$ $\frac{B _{a d ia}}{B _{i so}} = γ = \frac{C _{P}}{C _{V}}$