The translational kinetic energy of the molec | vedclass

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

Question

(D) The translational kinetic energy $(K)$ of a gas molecule is given by the formula: $K = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant a

Vedclass · Accepted Answer

increasing $T$ to $2T$

Vedclass · Answer

(D) The translational kinetic energy $(K)$ of a gas molecule is given by the formula: $K = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.From this relation, it is clear that $K \propto T$.If the kinetic energy is to be doubled, i.e.,$K' = 2K$, then the new temperature $T'$ must satisfy the relation $K' \propto T'$.Since $K' = 2K$, we have $T' = 2T$.Therefore, the absolute temperature must be increased to $2T$ to double the translational kinetic energy of the gas molecules.

The translational kinetic energy of the molecules of a gas at absolute temperature $T$ can be doubled by

Similar Questions

Mean kinetic energy (or average energy) per gram molecule of a monoatomic gas is given by

An ideal gas occupies a volume of $2\, m^3$ at a pressure of $3\times10^6\, Pa$. The energy of the gas is

The total kinetic energy of translatory motion of all the molecules of $5 \, L$ of nitrogen exerting a pressure $P$ is $3000 \, J$. Which of the following statements is correct?

$2\, kg$ of a monoatomic gas is at a pressure of $4\times10^4\, N/m^2$. The density of the gas is $8\, kg/m^3$. What is the order of energy of the gas due to its thermal motion?

At absolute zero temperature,the kinetic energy of the molecules

Vedclass Test Series

Exam Paper Generator

Online Exam Module