हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

The Figure Shows a Cylindrical Container Containing Oxygen (γ = 1.4) and Closed by a 50-kg Frictionless Piston.

Advertisements
Advertisements

प्रश्न

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.

संक्षेप में उत्तर
Advertisements

उत्तर

Given:
Mass of the piston (m) = 50 kg
Adiabatic constant of the gas, γ = 1.4
Area of cross-section of the piston (A) = 100 cm2
Atmospheric pressure (P0) = 100 kPa
g = 10 m/s2 
Distance moved by the piston , x = 20 cm
Work done by the gas,

dW=Pdv

The pressure (p) is because of two factors : the first is the initial pressure and 

Therefore,

`"W" = (("m""g")/"A" + "P"_0) xx "A""d"x`

`"W" = ((50 xx 10)/(100 xx 10^-4) + 10_5) xx 100 xx 10^-4 xx 20 xx 10^-4` 

W =( 5 × 104 +105) × 20 × 10-4

W = 1.5 × 105 × 20 × 10-4 

W = 300 J

Hence , nRdT = PΔV = 300

`=> "d""T" =300/("n""R")`

So, `"d""Q" = "n""C"_"p""d""T" = "n""c"_"p" xx (300/("n""R"))`

Using `"C"_"p" -"C"_"v" = "R" and "C"_"p"/"C"_"v" = gamma,`

`"d""Q" = ("n" gamma"R"300)/((gamma - 1) "n""R")`

`"d""Q" = ((300 xx 1.4 )/0.4) = 1050 "J"`

shaalaa.com
Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 27: Specific Heat Capacities of Gases - Exercises [पृष्ठ ७७]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 27 Specific Heat Capacities of Gases
Exercises | Q 3 | पृष्ठ ७७

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


Calculate the volume of 1 mole of an ideal gas at STP.


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


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


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.`


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.


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?


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.


An ideal gas of density 1.7 × 10−3 g cm−3 at a pressure of 1.5 × 105 Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities Cp and Cv of the gas.


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 ______.


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 ______.

  1. will be valid.
  2. will not be valid since the ions would experience forces other than due to collisions with the walls.
  3. will not be valid since collisions with walls would not be elastic.
  4. 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 ______.

  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

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×