At 27^{circ}C temperature,the kinetic energy | vedclass

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

Question

(A) The average kinetic energy $(KE)$ of an ideal gas is directly proportional to its absolute temperature $(T)$: $KE \propto T$.Given:Initial tempera

Vedclass · Accepted Answer

$2E_1$

Vedclass · Answer

(A) The average kinetic energy $(KE)$ of an ideal gas is directly proportional to its absolute temperature $(T)$: $KE \propto T$.Given:Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.Final temperature $T_2 = 327^{\circ}C = 327 + 273 = 600 \ K$.Initial kinetic energy = $E_1$.Using the proportionality relation:$\frac{E_1}{E_2} = \frac{T_1}{T_2}$Substituting the values:$\frac{E_1}{E_2} = \frac{300 \ K}{600 \ K} = \frac{1}{2}$Therefore,$E_2 = 2E_1$.

At $27^{\circ}C$ temperature,the kinetic energy of an ideal gas is $E_1$. If the temperature is increased to $327^{\circ}C$,then the kinetic energy would be:

Similar Questions

What is the pressure of hydrogen in a cylinder of volume $10 \text{ litre}$ if its total energy of translation is $7.5 \times 10^3 \text{ J}$?

$A$ gas has volume $V$ and pressure $P$. The total translational kinetic energy of all the molecules of the gas is

According to the kinetic theory of gases,the total energy of an ideal gas is equal to:

If $E$ is the kinetic energy per mole of an ideal gas and $T$ is the absolute temperature,then the universal gas constant is given as

By using which equation can the force acting on the walls of a container due to gas molecules be obtained?

Vedclass Test Series

Exam Paper Generator

Online Exam Module