MCQ
In the gas equation PV = RT, V is the volume of:
- A1 mol of gas
- B1g of gas
- CGas
- D1 litre of gas
✓
Answer
- 1 mol of gas
Explanation
For an Ideal Gas, PV = nRT
Here V is the volume of n moles of gas.
Thus for PV = (1)RT, V is the volume of 1 mol of gas.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
What would be the reason for the absence of atmosphere on the Moon?
The average degree of freedom per molecule of a gas is $6.$ The gas performs $25 \,J$ work, while expanding at constant pressure. The heat absorbed by the gas is ...... $J$
→→
Match the columns
→| Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
| $A.$ $1$ | $1.$ ${60^o}$ |
| $B.$ $4$ | $2.$ ${30^o}$ |
| $C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
| $D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |
A student says that he had applied a force $\text{F}=-\text{k}\sqrt{\text{x}}$ on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.
→- As x increases k increases.
- As x increases k decreases.
- As x increases k remains constant.
- The motion cannot be simple harmonic.
A tuning fork of frequency 512Hz is vibrated with a sonometer wire and 6 beats per second are heard. The beat frequency reduces if the tension in the string is slightly increased. The original frequency of vibration of the string is:
- 506Hz
- 512Hz
- 518Hz
- 524Hz.
The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is :
→Two sitar strings $A$ and $B$ playing the note $'Ga'$ are slightly out of tune and produce beats of frequency $6\,Hz$ . The tension in the string $A$ is slightly reduced and the beat frequency is found to reduce to $3\,Hz$. If the original frequency of $A$ is $324\,Hz$, what is the frequency of $B$ ..... $Hz$
→A closed pipe of length $10 \,cm$ has its fundamental frequency half that of the second overtone of an open pipe. The length of the open pipe ......... $cm$
→A hollow sphere has radius $6.4 \,m$. Minimum velocity required by a motor cyclist at bottom to complete the circle will be ........... $m/s$
→