The kinetic energy of translation of 20 , g o | vedclass

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

Question

(A) Given data:Mass of oxygen,$m = 20 \, g$Molar mass of oxygen,$M = 32 \, g/mol$Number of moles,$n = \frac{m}{M} = \frac{20}{32} = 0.625 \, mol$Tempe

Vedclass · Accepted Answer

$2490 \, J$

Vedclass · Answer

(A) Given data:Mass of oxygen,$m = 20 \, g$Molar mass of oxygen,$M = 32 \, g/mol$Number of moles,$n = \frac{m}{M} = \frac{20}{32} = 0.625 \, mol$Temperature,$T = 47^{\circ}C = 47 + 273 = 320 \, K$Gas constant,$R = 8.3 \, J/mol \cdot K$The kinetic energy of translation for an ideal gas is given by $E = \frac{3}{2} nRT$.Substituting the values:$E = \frac{3}{2} 	imes 0.625 	imes 8.3 	imes 320$$E = 1.5 	imes 0.625 	imes 8.3 	imes 320$$E = 2490 \, J$Thus,the kinetic energy of translation is $2490 \, J$.

The kinetic energy of translation of $20 \, g$ of oxygen at $47^{\circ}C$ is (molecular weight of oxygen is $32 \, g/mol$ and $R = 8.3 \, J/mol \cdot K$).

Similar Questions

If the temperature of a gas is increased from $27^{\circ}C$ to $927^{\circ}C$,what will be the change in its kinetic energy?

$A$ flask contains argon and oxygen in the ratio of $3: 2$ in mass and the mixture is kept at $27^{\circ} C$. The ratio of their average kinetic energy per molecule respectively will be ...........

The pressure exerted by an ideal gas at a particular temperature is directly proportional to

According to the Kinetic Theory of Gases,at a given temperature:

Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on the kinetic energy of molecules?

Vedclass Test Series

Exam Paper Generator

Online Exam Module