A container has N molecules at absolute tempe | vedclass

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

Question

(B) The total kinetic energy of an ideal gas is given by $U = \frac{3}{2} N k_B T$,where $N$ is the number of molecules,$k_B$ is the Boltzmann constan

Vedclass · Accepted Answer

$\frac{T}{2}$

Vedclass · Answer

(B) The total kinetic energy of an ideal gas is given by $U = \frac{3}{2} N k_B T$,where $N$ is the number of molecules,$k_B$ is the Boltzmann constant,and $T$ is the absolute temperature.Initially,the total kinetic energy is $U_i = \frac{3}{2} N k_B T$.Finally,the number of molecules becomes $N' = 2N$ and the total kinetic energy remains the same,so $U_f = U_i$.Let the new temperature be $T'$. Then $U_f = \frac{3}{2} (2N) k_B T'$.Equating the energies: $\frac{3}{2} N k_B T = \frac{3}{2} (2N) k_B T'$.Simplifying the equation: $T = 2T'$.Therefore,the new temperature is $T' = \frac{T}{2}$.

$A$ container has $N$ molecules at absolute temperature $T$. If the number of molecules is doubled but the total kinetic energy in the box remains the same as before,the absolute temperature of the gas is ...........

Similar Questions

If the temperature of a gas increases from $27^{\circ}C$ to $927^{\circ}C$,the $K.E.$ will be

If the average kinetic energy of a gas molecule at $27^{\circ} C$ is $3.3 \times 10^{-20} \,J$, then the average kinetic energy of the gas molecules at $127^{\circ} C$ is

Absolute zero temperature is the temperature at which:

Write the equation of pressure of an ideal gas in terms of its density.

The pressure of a gas is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module