Advertisements
Advertisements
प्रश्न
Calculate the volume of 1 mole of an ideal gas at STP.
Advertisements
उत्तर
Here,
STP means a system having a temperature of 273 K and 1 atm pressure.
Pressure, P = 1.01325\[\times\]05 Pa
No of moles, n = 1 mol
Temperature, T = 273 K
Applying the equation of an ideal gas, we get
PV = nRT
⇒ V =\[\frac{RT}{P}\]
⇒ V=\[\frac{8 . 314 \times 273}{1 . 01325 \times {10}^5} = 0 . 0224 \text{ m}^3\]
APPEARS IN
संबंधित प्रश्न
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
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 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 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 is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by `"C" ="C"_"v" +"R"/2.`
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.
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.
Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ = 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.
Two vessels A and B of equal volume V0 are connected by a narrow tube that can be closed by a valve. The vessels are fitted with pistons that can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (Cp/Cv = γ) at atmospheric pressure p0 and atmospheric temperature T0. The walls of vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperatures and pressures in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and pressure.
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 ______.
ABCDEFGH is a hollow cube made of an insulator (Figure). Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen. The usual kinetic theory expression for pressure ______.
- will be valid.
- will not be valid since the ions would experience forces other than due to collisions with the walls.
- will not be valid since collisions with walls would not be elastic.
- will not be valid because isotropy is lost.
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.
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).
