Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
For the first ideal gas,
Cp1 = specific heat at constant pressure
Cv1 = specific heat at constant volume
n1 = number of moles of the gas
`"C"_("p"1)/"C"_("v"1) = gamma and "C"_("p"1) -"C"_("v"1) = "R"`
`=> gamma "C"_("v"1) - "C"_("v"1) ="R"`
`=> "C"_("v"1)(gamma-1) ="R"`
`=> "C"_("v"1) = "R"/(gamma -1)`
`"C"_("p"1) = gamma "R"/((gamma-1))`
For the second ideal gas,
Cp2 = specific heat at constant pressure
Cv2 = specific heat at constant volume
n2 = number of moles of the gas
`"C"_("p"2)/"C"_("v"2) = gamma and "C"_("p"2) -"C"_("v"2) = "R"`
`=> gamma "C"_("v"2) - "C"_("v"2) ="R"`
`=> "C"_("v"2)(gamma-1) ="R"`
`=> "C"_("v"2) = "R"/(gamma -1)`
`"C"_("p"2) = gamma "R"/((gamma-1))`
Given:
n1 = n2 = 1 : 2
dU1 = nCv1dt
dU2= 2nCv2dT
When the gases are mixed,
nCv1dT + 2nCv2dT = 3nCvdT
`"C"_"v" = ("C"_("v"1) +2"C"_("v"2))/3`
`= (" R"/(gamma-1) +(2"R")/(gamma-1))/3`
`= (3"R")/((gamma-1)3) = "R"/(gamma-1)`
Hence, Cp / Cv in the mixture is γ.
APPEARS IN
संबंधित प्रश्न
The energy of a given sample of an ideal gas depends only on its
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
Consider the quantity \[\frac{MkT}{pV}\] of an ideal gas where M is the mass of the gas. It depends on the
Calculate the volume of 1 mole of an ideal gas at STP.
A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules.
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
A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol−1) is moving on a floor at a speed of 50 m s−1. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.
The figure shows a cylindrical container containing oxygen (γ = 1.4) and closed by a 50-kg frictionless piston. The area of cross-section is 100 cm2, atmospheric pressure is 100 kPa and g is 10 m s−2. The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.
The ratio of the molar heat capacities of an ideal gas is Cp/Cv = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K (a) keeping the pressure constant (b) keeping the volume constant and (c) adiaba
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.
An ideal gas at pressure 2.5 × 105 Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure (b) the final temperature and (c) the work done by the gas in the process. Take γ = 1.5
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.
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 ______.
1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (Figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time ______.
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.
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 |
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).
