Advertisements
Advertisements
Question
Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?
Options
Rms speed of a molecule
Density
Pressure
Average magnitude of momentum.
Advertisements
Solution
Pressure
Pressure of an ideal gas is given by PV = \[\frac{1}{3}mn u^2\] .
APPEARS IN
RELATED QUESTIONS
The energy of a given sample of an ideal gas depends only on its
The average momentum of a molecule in a sample of an ideal gas depends on
Calculate the volume of 1 mole of an ideal gas at STP.
Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.
Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.
(a) Q = 0
(b) W = 0
(c) Q = W
(d) Q ≠ W
A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1° C if 3.0 cal of heat is added to it. The gas may be
(a) helium
(b) argon
(c) oxygen
(d) carbon dioxide
An amount Q of heat is added to a monatomic ideal gas in a process in which the gas performs a work Q/2 on its surrounding. Find the molar heat capacity for the process
An ideal gas (Cp / Cv = γ) is taken through a process in which the pressure and the volume vary as p = aVb. Find the value of b for which the specific heat capacity in the process is zero.
Two ideal gases have the same value of Cp / Cv = γ. What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?
Half mole of an ideal gas (γ = 5/3) is taken through the cycle abcda, as shown in the figure. Take `"R" = 25/3"J""K"^-1 "mol"^-1 `. (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c) Find the amount of heat liberated in the processes cd and da.

The volume of an ideal gas (γ = 1.5) is changed adiabatically from 4.00 litres to 3.00 litres. Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.
Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.
1 litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.
A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 500 ms–1 in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground ______.
Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory `pV = 2/3` E, E is ______.
- the total energy per unit volume.
- only the translational part of energy because rotational energy is very small compared to the translational energy.
- only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum.
- the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero.
When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases ______.
- because of collisions with moving parts of the wall only.
- because of collisions with the entire wall.
- because the molecules gets accelerated in their motion inside the volume.
- because of redistribution of energy amongst the molecules.
The container shown in figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain µ1 = 4.0 and µ2 = 5.0 moles of a gas at pressures p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.
| V1 | V2 |
| µ1, p1 | µ2 |
| p2 |
