The kinetic energy of one gram molecule of a | vedclass

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

Question

(D) The kinetic energy of one mole (one gram molecule) of an ideal gas is given by the formula:$K = \frac{3}{2} RT$Given values are:$R = 8.31\,J/mol\c

Vedclass · Accepted Answer

$3.4 	imes 10^3\,J$

Vedclass · Answer

(D) The kinetic energy of one mole (one gram molecule) of an ideal gas is given by the formula:$K = \frac{3}{2} RT$Given values are:$R = 8.31\,J/mol\cdot K$$T = 273\,K$ (Standard temperature at $STP$)Substituting these values into the formula:$K = \frac{3}{2} 	imes 8.31 	imes 273$$K = 1.5 	imes 8.31 	imes 273$$K = 12.465 	imes 273$$K \approx 3402.945\,J$Rounding to the appropriate significant figures,we get:$K \approx 3.4 	imes 10^3\,J$

The kinetic energy of one gram molecule of a gas at normal temperature and pressure is $(R = 8.31\,J/mol\cdot K)$.

Similar Questions

What is the kinetic energy of $1$ mole of a monatomic gas?

What is the kinetic energy of $1$ mole of a monoatomic gas at $0 \, ^\circ C$? $(R = 8.31 \, J/mol \cdot K)$

Absolute zero temperature is the temperature at which:

Pressure of an ideal gas at constant volume is proportional to

At what temperature is the kinetic energy of a gas molecule double that of its value at $27^{\circ}C$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module