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 =B textadia=&gamma; p Isothermal bulk modulus =B text iso =p ⇒ (B text adia /B text iso )=&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 $= B_{adia} = γ p$
Isothermal bulk modulus $= B_{iso} = p$
$\Rightarrow \frac{B _{adia}}{B _{iso}} = γ = \frac{C _{p}}{C _{V}}$

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 $= B_{adia} = γ p$
Isothermal bulk modulus $= B_{iso} = p$
$\Rightarrow \frac{B _{adia}}{B _{iso}} = γ = \frac{C _{p}}{C _{V}}$