मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

Can Two States of an Ideal Gas Be Connected by an Isothermal Process as Well as an Adiabatic Process?

Advertisements
Advertisements

प्रश्न

Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?

थोडक्यात उत्तर
Advertisements

उत्तर

For two states to be connected by an isothermal process,

P1V1 = P2V2 ... (i)

For the same two states to be connected by an adiabatic process,

P1V1γ  = P2V2γ ...(ii)

If both the equations hold simultaneously then, on dividing eqaution (ii) by (i) we get

V1γ-1 = V2γ-1

Let the gas be monatomic. Then,

γ =`5/3`

`=> "V"_1^ (2/3)= "V"_2^(2/3)`

⇒ V1 = V2

If this condition is met, then the two states can be connected by an isothermal as well as an adiabatic process.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 27: Specific Heat Capacities of Gases - Short Answers [पृष्ठ ७६]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 27 Specific Heat Capacities of Gases
Short Answers | Q 9 | पृष्ठ ७६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?


The specific heat capacity of water is 


Can we define specific heat capacity at constant temperature?


Can we define specific heat capacity for an adiabatic process?


Does a solid also have two kinds of molar heat capacities Cp and Cv? If yes, is Cp > Cv? Or is Cp − Cv = R?


In a real gas, the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Examining the derivation of Cp − Cv = R, find whether Cp − Cv will be more than R, less than R or equal to R for a real gas.


Can a process on an ideal gas be both adiabatic and isothermal?


Show that the slope of the p−V diagram is greater for an adiabatic process compared to an isothermal process.


Two samples A and B are initially kept in the same state. Sample A is expanded through an adiabatic process and the sample B through an isothermal process. The final volumes of the samples are the same. The final pressures in A and B are pA and pBrespectively.


Let ∆Wa and ∆Wb be the work done by the systems A and B, respectively, in the previous question.


A sample of air weighing 1.18 g occupies 1.0 × 103 cm3 when kept at 300 K and 1.0 × 105 Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is  4.2 × 107 erg cal−1. Assume that air behaves as an ideal gas.


An ideal gas expands from 100 cm3 to 200 cm3 at a constant pressure of 2.0 × 105 Pa when 50 J of heat is supplied to it. Calculate (a) the change in internal energy of the gas (b) the number of moles in the gas if the initial temperature is 300 K (c) the molar heat capacity Cp at constant pressure and (d) the molar heat capacity Cv at constant volume.


In Joly's differential steam calorimeter, 3 g of an ideal gas is contained in a rigid closed sphere at 20°C. The sphere is heated by steam at 100°C and it is found that an extra 0.095 g of steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J g−1 K−1. The latent heat of vaporisation of water = 540 cal g−1 


An engine takes in 5 moles of air at 20°C and 1 atm, and compresses it adiabatically to `1/10^"th"` of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is ______.


If at same temperature and pressure, the densities for two diatomic gases are respectively d1 and d2 then the ratio of velocities of sound in these gases will be ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×