CO₂(O -C-O )is a triatomic gas. Mean kinetic energy of one gram gas will be (If N-Avogadro's number, k-Boltzmann's constant and molecular weight of CO₂ = 44)

The correct option is 7/88Nkt.

Let's derive the mean kinetic energy of one gram of carbon dioxide (${\mathrm{CO}}_2$) gas step-by-step:

Step 1: Understand the degrees of freedom of the gas molecule
Carbon dioxide ($O-C-O$) is a linear triatomic gas molecule. A linear triatomic molecule has:
- Translational degrees of freedom = 3
- Rotational degrees of freedom = 2
- Vibrational degrees of freedom = 2 (at normal temperatures, active vibrational modes contribute to the degrees of freedom, giving a total effective degrees of freedom, $f=7$).

Step 2: Calculate the mean kinetic energy per molecule
According to the law of equipartition of energy, the average kinetic energy associated with each degree of freedom per molecule is:

$E=\frac{1}{2}kt$

Therefore, the total mean kinetic energy of one molecule with $f=7$ degrees of freedom is:

$E_{\text{molecule}}=\frac{7}{2}kt$

where $k$ is the Boltzmann's constant and $t$ is the absolute temperature.

Step 3: Calculate the number of molecules in 1 gram of the gas
The molecular weight of ${\mathrm{CO}}_2$ is given as 44 g/mol.
Since 1 mole of any gas contains Avogadro's number ($N$) of molecules:
- Number of molecules in 44 grams of ${\mathrm{CO}}_2$ = $N$
- Number of molecules in 1 gram of ${\mathrm{CO}}_2$ is:

$N_{\text{1g}}=\frac{N}{44}$

Step 4: Find the total mean kinetic energy of 1 gram of the gas
Multiply the number of molecules in 1 gram by the mean kinetic energy of a single molecule:

$E_{\text{total}}=N_{\text{1g}}\times E_{\text{molecule}}$

$E_{\text{total}}=\left(\frac{N}{44}\right)\times \left(\frac{7}{2}kt\right)$

$E_{\text{total}}=\frac{7}{88}Nkt$

Thus, the mean kinetic energy of one gram of the gas is indeed $\frac{7}{88}Nkt$.