Advertisements
Advertisements
प्रश्न
We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state?
(Hydrogen molecules can be consider as spheres of radius 1 Å).
उत्तर
Assuming hydrogen molecules as spheres of radius 1 Å.
So, r = 1 Å = radius
The volume of hydrogen molecules = `4/3 pir^3`
= `4/3 (3.14)(10^-10)^3`
= `4 xx 10^-30 m^3`
Number of moles of H2 = `"Mass"/"Molecular mass"`
= `0.5/2`
= 0.25
Molecules of H2 present = Number of moles of H2 present × 6.023 × 1023
= 0.25 × 6.023 × 1023
∴ Volume of molecules present = Molecules number × Volume of each molecule
= 0.25 × 6.023 × 1023 × 4 × 10–30
= 6.023 × 1023 × 10–30
= 6 × 10–7 m3 ......(i)
Now, if the ideal gas law is considered to be constant,
`p_iV_i = p_fV_f`
`V_f = (p_i/p_f)`
`V_i = (1/100)(3 xx 10^-2)^3`
= `(27 xx 10^-6)/10^2`
= 2.7 × 10–7 m3 ......(ii)
Hence, on compression, the volume of the gas is of the order of the molecular volume [form equation (i) and equation (ii)]. The intermolecular forces will play a role and the gas will deviate from ideal gas behaviour.
APPEARS IN
संबंधित प्रश्न
Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?
The average momentum of a molecule in a sample of an ideal gas depends on
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
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.
The figure shows an adiabatic cylindrical tube of volume V0 divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p1 and temperature T1 is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part. Cp/Cv = γ is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find (a) the volumes of the two parts (b) the heat given to the gas in the left part and (c) the final common pressure of the gases.
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.
In a diatomic molecule, the rotational energy at a given temperature ______.
- obeys Maxwell’s distribution.
- have the same value for all molecules.
- equals the translational kinetic energy for each molecule.
- is (2/3)rd the translational kinetic energy for each molecule.
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 |
