English
Karnataka Board PUCPUC Science Class 11

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? - Physics

Advertisements
Advertisements

Question

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?

Answer in Brief
Advertisements

Solution

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.

shaalaa.com
Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
  Is there an error in this question or solution?
Chapter 5: Specific Heat Capacities of Gases - Exercises [Page 78]

APPEARS IN

HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 5 Specific Heat Capacities of Gases
Exercises | Q 11 | Page 78

RELATED QUESTIONS

The energy of a given sample of an ideal gas depends only on its


Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?


Find the number of molecules in 1 cm3 of an ideal gas at 0°C and at a pressure of 10−5mm of mercury.

Use R = 8.31 J K-1 mol-1


Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.


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


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.


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.


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.


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


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.


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 ______.


In a diatomic molecule, the rotational energy at a given temperature ______.

  1. obeys Maxwell’s distribution.
  2. have the same value for all molecules.
  3. equals the translational kinetic energy for each molecule.
  4. 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 ______.

  1. because of collisions with moving parts of the wall only.
  2. because of collisions with the entire wall.
  3. because the molecules gets accelerated in their motion inside the volume.
  4. 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

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).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×