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Van der Waal's equation

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  Van der Waal's equation | 1st year chemistry unit 3 Why the volume correction is done by van der Waal? Ans. Actual volume of gas molecules is not negligible as compared to the total volume of vessel. Therefore, Van der Waal corrected the volume factor in general gas equation. So, excluded volume is subtracted from total volume of vessel to get the free volume available for gas molecules.                                       V free = V vessel – b (excluded volume) Why the pressure correction is done by Van der Waal? Ans. Forces of attraction present between molecules of real gas. Therefore, Van der Waal corrected the pressure correction in general gas equation. Lessened pressure is subtracted from ideal pressure to get the observed pressure.        ...

Non ideal behaviour of gases

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  Non ideal behaviour of gases | 1st year chemistry chapter 3 Gases deviate more from the general gas equation at 0 o C and deviate to less extent at 100 o C. Why? Ans. General gas equation is for ideal gases (which have no forces of attraction). At 0 o C molecules are close to each other and have strong attraction, while at 100 o C molecules are far away from each other and have very weak attractions. Therefore, Gases deviate more from the general gas equation at 0 o C and deviate to less extent at 100 o C. Hydrogen and helium are ideal at room temperature, but SO 2 and Cl 2 are non-ideal. How do you explain it? Ans. Molecules of hydrogen and helium are very far away and have no force of attraction at room temperature. While molecules of SO 2 and Cl 2 are very close and have force of attraction at room temperature. Therefore, Hydrogen and helium are ideal at room temperature, but SO 2 and Cl 2 are non-ideal. Why the gases deviate from ideal behaviour at high pressure ...

Liquefaction of gases

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  Liquefaction of gases | first year chemistry chapter 3 Define critical temperature. Give examples. Ans. The temperature of a gas above which a gas cannot be liquefied, no matter how great the  pressure is applied, is called  critical temperature. It is denoted by T c . Examples: Gas Critical temperature ( o C) O 2 -118.7 NH 3 132.4 CO 2 31.1 Define critical pressure. Give examples. Ans. It is the minimum pressure which is required to liquefy a gas at its critical temperature. It is denoted by P c . Examples:  Gas Critical pressure (atm) O 2 49.7 NH 3 111.5 CO 2 73.0 Define critical volume. Give examples. Ans. The volume which is occupied by one mole of a gas at critical temperature and critical pressure is called critical volume. It is denoted by V c. Examples: The critical volu...

Graham's law of diffusion

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  Graham's law of diffusion | chemistry class 11 | fsc part 1 | chemistry notes Q) Define diffusion. Give examples. Ans. The spontaneous mixing of the molecules of different gases by random motion and collisions to form a homogeneous mixture is called diffusion. Examples:     i) The spreading of fragrance of rose in air                    ii) The gases of balloon diffuse in air after its burst.   Q) Define effusion. Give examples. Ans. The escaping of gas molecules one by one through a tiny hole into an evacuated space is called effusion. Example: An inflated balloon gradually becomes smaller in size due to its effusion.   Q) State Graham’s law of diffusion. Write its mathematical form. Ans. The rate of diffusion or effusion of a gas is inversely proportional to the square root of its density at constant temperature and pressure. Mathematical form...

Dalton law of partial pressure | 1st year chemistry chapter 3 short questions notes | dalton law of partial pressure class 11

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  Dalton law of partial pressure. 1st year chemistry chapter 3 short questions notes. Dalton law of partial pressure class 11   Q) State Dalton's law of partial pressure. Why is it necessary that no forces of attractions should be there to obey this law?   Ans. Statement : Total pressure exerted by mixture of gases is equal to sum of individual partial  pressures of gases at given temperature.  Mathematically:    P t = P 1 + P 2 + P 3 If forces of attraction present between molecules of gases then total pressure will be less than individual partial pressure of gases. Therefore it is necessary that no forces of attraction should be there to obey this law.   Q) Define Mole fraction of gas. Ans. It is the ratio of number of moles of a gas to the total number of moles of gases present in a mixture. Its value always smaller than 1. Q) What is partial pressure? Give its formula. Ans. Partial pressure is the pressure which is exerted by a gas...

avogadro's law of gases | 1st year chemistry chapter 3 important short questions

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 A vogadro's law of gases . 1st year chemistry chapter 3 important short questions To watch lecture video of this topic on youtube please   click here What is Avogadro’s law of gases? Ans. Equal volume of all ideal gases contain equal number of molecules at S.T.P. Justify that 1 cm 3 of H 2 and 1 cm 3 of CH 4 at STP will have same number of molecules. When one molecule of CH 4 is 8 times heavier than that of hydrogen.   Ans. According to Avogadro’s law “ Equal volume of all ideal gases contain equal number of molecules at S.T.P.” 1 cm 3 of H 2 and 1 cm 3 of CH 4 have same volume so these have same number of molecules. No doubt CH 4 is 8 times heavier than that of hydrogen, but volume not depends upon sizes and masses of molecules.   Related searches of this topic: avogadro law state and explain avogadro's law define avogadro's law avogadro law class 11 example avogadro's law avogadro law class 11 avogadro law class 11 chemistry avogadro law che...

General gas equation | 1st year chemistry chapter 3 important short questions | first year chemistry notes

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 General gas equation | 1st year chemistry chapter 3 important short questions | first year chemistry notes To watch lecture video of this topic on youtube please   click here How the value of the general gas constant ‘R’ can be derived with the help of Avogadro’s law? Ans. According to Avogadro’s law, the volume of one mole of all the ideal gases at S.T.P are 22.414 dm 3 . Putting the values of                 P = 1 atm                                           T = 273.16 K                                           V = 22.4 dm 3                                            n = 1 ...