The kinetic energy of translation of the mole | vedclass

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

Question

(B) The translational kinetic energy of $N$ molecules is given by $K = \frac{3}{2} N k_B T = \frac{3}{2} n R T$,where $n$ is the number of moles.Given

Vedclass · Accepted Answer

$4102.8$

Vedclass · Answer

(B) The translational kinetic energy of $N$ molecules is given by $K = \frac{3}{2} N k_B T = \frac{3}{2} n R T$,where $n$ is the number of moles.Given mass $m = 50 	ext{ g}$,molar mass of $	ext{CO}_2$ $M = 44 	ext{ g/mol}$.Number of moles $n = \frac{m}{M} = \frac{50}{44} \approx 1.136 	ext{ mol}$.Temperature $T = 17 + 273.15 = 290.15 	ext{ K}$.Using $R = 8.314 	ext{ J/(mol K)}$:$K = \frac{3}{2} 	imes \left( \frac{50}{44} ight) 	imes 8.314 	imes 290.15$.$K = 1.5 	imes 1.13636 	imes 8.314 	imes 290.15 \approx 4108.6 	ext{ J}$.Comparing with the given options,the closest value is $4102.8 	ext{ J}$.

The kinetic energy of translation of the molecules in $50 \text{ g}$ of $\text{CO}_2$ gas at $17^{\circ}\text{C}$ is: (in $\text{ J}$)

Similar Questions

The average kinetic energy per gram of a monoatomic gas is given by:

Consider a gas with $n$ molecules. The average number of molecules moving in a fixed direction is ..........

$A$ gas is contained in a vessel at pressure $P_0$. If the mass of all molecules is halved and their speed is doubled,what will be the final pressure?

Assuming the expression for the pressure exerted by the gas,it can be shown that pressure is

According to the kinetic theory of gases,the average kinetic energy of $1$ gram-mole of a gas is given by which of the following expressions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module