Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given:
Number of moles of the gas,
n = 0.5 mol
`"R" = 25/3 `J/mol -K
`gamma =5/3`
(a) Temperature at a = Ta
PaVa =nRTa
`=> "T""a" = ("P"_"a""V"_"a")/("n""R") = (100 xx 10^3 xx 5000 xx 10^-6)/(0.5 xx 25/3) = 120 "K"`
Similarly, temperature at b,
`"T"_"b" = ("P"_b"V"_"b")/("n""R")`
`"T"_"b" =( 100 xx 10^3 xx 10000 xx 10^-6)/(0.5 xx 25/3)`
Tb =240 K
Similarly, temperature at c is 480 K and at d is 240 K.
(b) For process ab,
dQ = ncpdT
[Since ab is isobaric]
`"d""Q" = 1/2 xx ("R"gamma)/(gamma-1) ("T"_"b" - "T"_"a")`
`"d""Q" =1/2 xx ((25 xx 5)/(3 xx 3))/(5/3 -1 ) xx (240 -120 )`
`"d""Q" = 1/2 xx 125 /9 xx 3/2 xx (120)`
dQ = 1250 J
For line bc, volume is constant. So, it is an isochoric process.
dQ = dU + dW
[dW = 0, isochoric process]
dQ = dU = nCvdT
dQ = nCv (Tc - Tb)
`"d""Q" = 1/2 xx ((25/3))/[[(5/3)-1]] xx (240)`
`"d""Q" = 1/2 xx 25/3 xx 3/2 xx 240 =1500 "J"`
(c) Heat liberated in cd (isobaric process),
dQ = − nCpdT
`"d""Q" = -1/2 xx( gamma "R")/(gamma-1) xx ("T"_d -"T"_"c")`
`"d""Q" = -1/2 xx 125/9 xx 3/2 xx ( 240 -480)`
`"d""Q" = -1/2 xx 125/6 xx 240 = 2500 "J"`
Heat liberated in da (isochoric process),
⇒ dQ = dU
Q= −nCvdT
`"d""Q" = -1/2 xx "R" /(gamma -1 )("T"_"a"-"T"_"d")`
`"d""Q" = -1/2 xx 25/2 xx (120-240)`
`"d""Q" = 25/4 xx 120 =750 "J"`
APPEARS IN
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
Calculate the volume of 1 mole of an ideal gas at STP.
A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules.
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.
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.
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.
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.
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.
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 ______.
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.
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.
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).
