Advertisements
Advertisements
प्रश्न
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.
उत्तर
For an adiabatic process, PVγ = Constant
So, P1V1γ = P2V2γ ...(i)
According to the problem,
V1 + V2 = V0 ...(ii)
Using the relation in eq (ii) in eq (i), we get
P1V1γ = P2(V0 − V1)γ
Or `("P"_1/"P"_2)^(1/gamma) = ("V"_0 -"V"_1)/"V"_1`
`"V"_1"P"_1^(1/gamma) = "V"_0"P"_2^(1/gamma) - "V"_1"P"_2^(1/gamma)`
`"V"_1( "P"_1^(1/gamma) + "P"_2^(1/gamma)) = "V"_0"P"_2^(1/gamma)`
`"V"_1 = ("P"_2^(1-gamma) "V"_0)/("P"_1^(1/gamma) +"P"_2^(1/gamma))`
Using equation (ii), we get
`"V"_2 = ("P"_1^(1/gamma) "V"_0)/("P"_1 ^(1/gamma)+ "P"_2^(1/gamma))`
(b) Since the whole process takes place in adiabatic surroundings, the separator is adiabatic.
Hence, heat given to the gas in the left part is 0.
(c) There will be a common pressure 'P' when equilibrium is reached. The slid will move until the pressure on the two sides becomes equal.
P1V1γ + P2V2γ = PV0γ
For equilibrium, `"V"_1 = "V"_2 = "V"_0/2`
Hence,
`"P"_1("V"_0/2)^gamma + "P"_2("V"_0/2) ^gamma = "P"("V"_0)^gamma`
Or `"P" = (("P" _1^(1/gamma) + "P"_2^(1/gamma))/2)^gamma`
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?
Consider the quantity \[\frac{MkT}{pV}\] of an ideal gas where M is the mass of the gas. It depends on the
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
A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules.
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
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 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.
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
Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.
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.
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.
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.
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).
