The kinetic energy of an ideal gas is E_0 at | vedclass

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

Question

(C) The kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.$K.E. \propto T$Given, initial te

Vedclass · Accepted Answer

$3E_0/2$

Vedclass · Answer

(C) The kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.$K.E. \propto T$Given, initial temperature $T_1 = 27^\circ$$C$ = $27 + 273 = 300$ $K$.Initial kinetic energy $K_1 = E_0$.Final temperature $T_2 = 177^\circ$$C$ = $177 + 273 = 450$ $K$.Using the relation $\frac{K_2}{K_1} = \frac{T_2}{T_1}$:$\frac{K_2}{E_0} = \frac{450}{300}$$\frac{K_2}{E_0} = \frac{3}{2}$$K_2 = \frac{3}{2} E_0$.

The kinetic energy of an ideal gas is $E_0$ at $27^\circ$$C$. When the temperature is increased to $177^\circ$$C$, then the kinetic energy will be

Similar Questions

If the distance between molecules is doubled at a constant temperature,the pressure will become .......

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

$A$ cylinder of capacity $200 \ L$ is filled with $H_2$ gas. The total average translational kinetic energy of the molecules is $1.52 \times 10^5 \ J$. Calculate the pressure of $H_2$ in the cylinder in $N \ m^{-2}$.

The energy (in $eV$) possessed by a neon atom at $27^{\circ} C$ is ..............

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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module