The ratio of the adiabatic to isothermal elasticities of a triatomic gas is

The correct option is 4/3.

To find the ratio of the adiabatic elasticity to the isothermal elasticity of a triatomic gas, we first determine the mathematical expressions for both elasticities.

Elasticity (or bulk modulus) is defined as the ratio of volume stress to volume strain:
$E=-\frac{dP}{dV/V}=-V\frac{dP}{dV}$

1. Isothermal Elasticity ($E_T$):
For an isothermal process, the temperature remains constant. The equation of state is Boyle's Law:
$PV=\text{constant}$

Differentiating both sides with respect to volume:
$PdV+VdP=0$

Rearranging the terms yields:
$-V\frac{dP}{dV}=P$

Thus, the isothermal elasticity is equal to the pressure of the gas:
$E_T=P$

2. Adiabatic Elasticity ($E_S$):
For an adiabatic process, there is no heat exchange with the surroundings. The equation of state is:
$P{V}^{\gamma }=\text{constant}$

Differentiating both sides:
$V^{\gamma }dP+\gamma P{V}^{\gamma -1}dV=0$

Dividing the equation by $V^{\gamma -1}$ gives:
$VdP+\gamma PdV=0$

Rearranging the terms yields:
$-V\frac{dP}{dV}=\gamma P$

Thus, the adiabatic elasticity is:
$E_S=\gamma P$

3. Ratio of Elasticities:
The ratio of the adiabatic elasticity to the isothermal elasticity is:
$\frac{E_S}{E_T}=\frac{\gamma P}{P}=\gamma$

Here, $\gamma$ is the ratio of specific heats ($C_p/C_v$). For a non-linear triatomic gas, the number of degrees of freedom ($f$) is 6.

The relation between $\gamma$ and the degrees of freedom $f$ is:
$\gamma =1+\frac{2}{f}$

Substituting $f=6$ into the equation:
$\gamma =1+\frac{2}{6}=1+\frac{1}{3}=\frac{4}{3}$

Therefore, the ratio of the adiabatic to isothermal elasticities of the triatomic gas is 4/3.