Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
(a) For line ab, volume is constant.
So, from the ideal gas equation,
`"P"_1/"T"_1 = "P"_2/"T"_2`
`=> 100/300 = 200 /"T"_2`
`"T"_2 = ((200 xx 300 )/100) = 600 "K"`
For line bc, pressure is constant.
`So, "V"_1 /"T"_1 = "V"_2 /"T"_2`
`=> 100/600 = 150 / "T"_2`
`=> "T"_2 ((600 xx 150) /100) = 900 "K"`
(b)
As process ab is isochoric, Wab =0 .
During process bc,
P = 200 kPa
The volume is changing from 100 to 150 cm3 .
Therefore, work done = 50 × 10−6 × 200 × 103 J
= 10 J
(c) For ab (isochoric process), work done = 0.
From the first law,
dQ = dU = nCvdT
⇒ Heat supplied = nCvdT
Now,
`"Q"_"ab" = (("P""V")/("R""T")) xx (("R")/(gamma - 1)) xx "d""T"`
`= (200 xx 10 ^ 3 xx 100 xx 10 ^-6 xx 300)/(600 xx 0.67)`
= 14.925 ( ∴ γ = 1.67)
F or bc (isobaric process) :
Heat supplied in bc = `"n" "C"_"p""d""T" ( "C"_"p" = (gamma "R")/(gamma - 1))`
`= ("P""V")/("R""T") xx (gamma "R") /(gamma -1 ) xx "d""T"`
`= (200 xx 10^3 xx 150 xx 10 ^-6 )/(600 xx 0.67) xx 300 `
`= 10 xx 1.67 /0.67 = 16.7/0.67 = 24.925`
(d) dQ = dU + dW
Now,
dU = dQ − dW
= Heat supplied − Work done
= (24.925 + 14.925) − 10
= 39.850 − 10 = 29.850 J.
APPEARS IN
संबंधित प्रश्न
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.
Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.
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 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.
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?
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.
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.
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.
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).
