Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Initial pressure of the gas in both the vessels = P0
Initial temperature of the gas in both the vessels = T0
Initial volume = V0
`"C"_"p"/"C"_"v" = gamma`
(a) The temperature inside the diathermic vessel remains constant. Thus,
P1V1 = P2V2
⇒ P0V0 = P2 × 2V0
`=> "P"_2 = "P"_0/2`
emperature = T0
The temperature inside the adiabatic vessel does not remain constant.
So, for an adiabatic process,
T1V1γ−1 = T2V2γ−1
⇒T0V0γ−1 = T × (2V0)γ−1
⇒ T2 = T0 × 21−γ
⇒ P1V1γ = P2V2γ
P0V0γ = P2 × (2V0)γ
`=> "P"_2 = ("P"_0/2^gamma)`
(b) When the valves are open, the temperature remains T0 throughout, i.e. T1 = T2 = T0 .
Also, pressure will be same throughout. Thus, P1 = P2. As, temperature has not changed on side 1, so pressure on this side will also not change (volume is also fixed due to fixed piston) and will be equal to P0 (pressure is an intrinsic variable).
On side 2, pressure will change to accommodate the changes in temperature on this side.
So, P0 = P1 + P2
⇒ 2P1 = 2P2
`"So", "P"_1 = "P"_2 ="P"_0/2`
APPEARS IN
संबंधित प्रश्न
Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?
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
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
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.
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.`
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.
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.
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.
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.
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.
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 ______.
- because of collisions with moving parts of the wall only.
- because of collisions with the entire wall.
- because the molecules gets accelerated in their motion inside the volume.
- because of redistribution of energy amongst the molecules.
