मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

The Volume of an Ideal Gas (γ = 1.5) is Changed Adiabatically from 4.00 Litres to 3.00 Litres. - Physics

Advertisements
Advertisements

प्रश्न

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.

थोडक्यात उत्तर

उत्तर

Given,
γ = 1.5
Since the process is adiabatic, PVγ = constant.

(a) P1V1γ = P2V2γ
Given, V1 = 4 L
           V2 = 3 L

we need to find `"P"_2/"P" _1.`

`=> "P"_2/"P"_1 = ("V"_1/"V"_2)`

` => (4/3)^1.5  = 1.5396 = 1.54`

(b) Also, for an adiabatic process,
TVγ−1 = constant
T1V1γ−1 = T2V2γ−1

`=> "T"_2/ "T"_1 = ("V"_1/"V"_2)^ (gamma-1) =(4/3)^0.5 = 1.154`

shaalaa.com
Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Specific Heat Capacities of Gases - Exercises [पृष्ठ ७८]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
पाठ 5 Specific Heat Capacities of Gases
Exercises | Q 16 | पृष्ठ ७८

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


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


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 (γ = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate (a) the temperatures at b and c (b) the work done in the process (c) the amount of heat supplied in the path ab and in the path bcand (d) the change in the internal energy of the gas in the process.


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


Consider a given sample of an ideal gas (Cp/Cv = γ) having initial pressure p0 and volume V0. (a) The gas is  isothermally taken to a pressure p0/2 and from there, adiabatically to a pressure p0/4. Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure p0/2 and from there, isothermally to a pressure p0/4. Find the final volume.


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.


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 ______.

  1. the total energy per unit volume.
  2. only the translational part of energy because rotational energy is very small compared to the translational energy.
  3. only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum.
  4. 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.

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×