The total kinetic energy of 0.6 mol of an id | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The total kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Given: $n = 0.6 \ mol$,$T = 27 \ ^oC

Vedclass · Accepted Answer

$2245$

Vedclass · Answer

(B) The total kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by the formula: $KE = \frac{3}{2} nRT$. Given: $n = 0.6 \ mol$,$T = 27 \ ^oC = 27 + 273 = 300 \ K$,and $R = 8.314 \ J \ mol^{-1} K^{-1}$. Substituting the values: $KE = \frac{3}{2} 	imes 0.6 	imes 8.314 	imes 300$. $KE = 1.5 	imes 0.6 	imes 8.314 	imes 300 = 0.9 	imes 8.314 	imes 300 = 2244.78 \ J$. Rounding to the nearest integer,we get $2245 \ J$.

The total kinetic energy of $0.6 \ mol$ of an ideal gas at $27 \ ^oC$ is .......... $J$.

Similar Questions

What is the internal energy of one mole of an ideal gas?

Given that the atomic weights of carbon,nitrogen,and oxygen are $12$,$14$,and $16$ respectively,which of the following pairs of gases will diffuse at the same rate?

For a fixed number of moles of a gas at constant volume,the pressure of the gas increases with an increase in temperature because....

At the same temperature and pressure,which of the following gases will have the highest kinetic energy $(K.E)$ per mole?

For a fixed number of moles of a gas at constant volume,the pressure of the gas increases with an increase in temperature because:

Vedclass Test Series

Exam Paper Generator

Online Exam Module