Question

calculate the most probable speed of co2 gas molecules at 27 degree c.

Answer 1

Answer:- $\frac{238m}{s}$ .

Solution:- velocity of the gas at a particular temperature possessed by the maximum fraction of the total number of molecules of the gas is known as most probable speed. It is represented by $\alpha$ and the formula to calculate this is:

$\alpha =\sqrt{\frac{2RT}{M}}$

where, R is the universal gas constant and it's value is 8.314 joule per mol per kelvin. T is the kelvin temperature and M is the molecular weight of the gas.

Given temperature is 27 degree C, let's convert it to kelvin:

27+273 = 300 K

M of carbon dioxide = 12.01+2(16.00) = 44.01 gram per mol

We need to take the molar mass in kg per mol.

$\frac{44.01g}{mol}(\frac{1kg}{1000g})$

= 0.04401 kg per mol

Also, we know that, $J=kg.m^2.s^-^2$

So, the value of R is $kg.m^2.s^-^2.mol^-^1.K^-^1$

Let's plug in the values in the formula:

$\alpha =\sqrt{\frac{8.314kg.m^2.s^-^2.mol^-^1.K^-^1(300K)}{0.04401kg.mol^-^1}}$

$\alpha =\frac{238m}{s}$

So, the most probable speed of carbon dioxide gas molecules at 27 degree C is $\frac{238m}{s}$ .