If the potential energy of a gas molecule is $U = \frac{M}{r^6} - \frac{N}{r^{12}}$,where $M$ and $N$ are positive constants,then the potential energy at equilibrium must be

  • A
    Zero
  • B
    $M^2 / 4N$
  • C
    $NM^2 / 4$
  • D
    $MN^2 / 4$

Explore More

Similar Questions

$A$ gas is enclosed in a closed pot. On keeping this pot in a train moving with high speed,the temperature of the gas

Initially,a gas of diatomic molecules is contained in a cylinder of volume $V_{1}$ at a pressure $P_{1}$ and temperature $250\, K$. Assuming that $25\%$ of the molecules get dissociated,causing a change in the number of moles. The pressure of the resulting gas at temperature $2000\, K$,when contained in a volume $2V_{1}$,is given by $P_{2}$. The ratio $\frac{P_{2}}{P_{1}}$ is:

Match Column $- I$ and Column $- II$ and choose the correct match from the given choices.
Column $- I$Column $- II$
$(A)$ Root mean square speed of gas molecules$(P)$ $\frac{1}{3} n m \bar{v}^{2}$
$(B)$ Pressure exerted by ideal gas$(Q)$ $\sqrt{\frac{3 RT}{M}}$
$(C)$ Average kinetic energy of a molecule$(R)$ $\frac{5}{2} RT$
$(D)$ Total internal energy of $1$ mole of a diatomic gas$(S)$ $\frac{3}{2} k_{B} T$

$A$ cylinder of fixed capacity $44.8 \, L$ contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0 \, ^{\circ}C$? $(R = 8.31 \, J \, mol^{-1} K^{-1})$

Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same, are compressed till their volume becomes half the initial volume.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo