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D Divya Prakash Singh
The equation of van der Waals after taking into account the corrections for pressure and volume, Where a and b are called van der Waals constants or parameters. Here the significance of a and b is important: Value of 'a' is a measure of the magnitude of intermolecular attractive forces within the gas and is independent of temperature and pressure. And the value of 'b' is the volume occupied by...

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D Divya Prakash Singh
The critical temperature we know is the highest temperature at which liquid exists above it is gas. Given that the critical temperature of Carbon dioxide and Methane are.  Higher is the critical temperature of a gas, easier is its liquefaction. So, as the critical temperature increases the gas is now easier to liquefaction. That means the intermolecular forces of attraction between the...

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D Divya Prakash Singh
Charles' Law states that pressure remaining constant, the volume of a fixed mass of a gas is directly proportional to its absolute temperature. Charles found that for all gases, at any given pressure, the graph of volume vs temperature (in Celsius) is a straight line and on extending to zero volume, each line intercepts the temperature axis at – 273.15 ° C. We can see that the volume of the...

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D Divya Prakash Singh
Given quantity, , The SI units of each factors are, For pressure p is . For Volume V is . For Temperaute T is. For number of moles, n is mol. Therefore we have for the quantity  The SI unit is 

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D Divya Prakash Singh
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. So, let us assume the weight of dihydrogen in a total mixture weight of  be  and dioxygen be . Then, the number of moles of  will be,  and the number of moles of , . And given that the total pressure of the mixture is . Then we have a partial pressure of , Therefore, the partial pressure of  is .

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D Divya Prakash Singh
Let the molar mass of the gas be , Then, given that  of a gas at  occupied the same volume as   of  at  So, we have the relation,    and  Therefore  wihch gives,  Or,  Or,   

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D Divya Prakash Singh
Given the mass of carbon dioxide is 8.8grams at  and at 1 bar pressure. So, the number of moles of  Now, the pressure of  Given  also, the The Ideal gas equation;  So volume occupied by 8.8 g of carbon dioxide is 5.048L

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D Divya Prakash Singh
The payload can be defined as: (Mass of the displaced air - Mass of the balloon) Given the radius of the balloon, r = 10 m. Mass of the balloon, m = 100 kg. Therefore, the volume of the balloon will be: Now, the volume of the air displaced: The mass of the air displaced :         Let  be the mass of helium gas filled into the balloon, then Or,   approximately. The balloon is filled with He...

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D Divya Prakash Singh
Given the mass of oxygen gas and mass of hydrogen gas. Molar mass of  Therefore  Molar mass of  Therefore   Total number of moles of mixture  and  So, Ideal gas equation;  or  Therefore Total pressure is 

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D Divya Prakash Singh
Given that  grains are distributed each second. Time taken to distribute  grains = 1second. Time taken to distribute one Avogadro number of wheat grains.

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D Divya Prakash Singh
Dinitrogen   has a molar mass . Given 1.4g of  gas. That means   number of molecules. And as 1 molecule of  has 14 electrons. So,  molecules of  has  electrons.

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D Divya Prakash Singh
We have the ideal gas equation,   Pressure, p=3.32bar Volume, V=5 Number of moles, n=4 mol Gas constant R= 0.083 Temperature,  T=? so, temperature, T=50K

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D Divya Prakash Singh
Assume the round-bottomed flask has volume  so, the volume of air in the flask at  . To find out how much air has been expelled out, we use Charle's Law: Therefore  Or, Fraction of air expelled     

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D Divya Prakash Singh
Given, The volume of phosphorus vapour   which weights about  . Temperature  Pressure Hence we apply the ideal gas equation,  , where the number of moles will be   Therefore the molar mass is 

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D Divya Prakash Singh
Given the density of a gas is equal to  at  and at 2bar pressure. Density , for the same gas at different temperatures and pressures we can apply, Here,  , ,  then at STP, we have      or    

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D Divya Prakash Singh
Pressure of the gas mixture will be the sum of partial pressure of  gas and partial pressure of  gas. So, calculating the partial pressures of each gas.  Partial pressure of  in a 1L volume vessel. ,  As the temperature remains constant,  or     or    Now, calculating the partial pressure of  gas in 1L vessel.   or   Therefore the total pressure  

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D Divya Prakash Singh
Given a mixture of 3.2g of methane and 4.4g of carbon dioxide contained a  flask at . So, the pressure exerted by the mixture will be The pressure exerted by the Methane gas, The pressure exerted by the Carbon dioxide gas, So, the total pressure exerted  And in terms of SI units, we get,

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D Divya Prakash Singh
We have the chemical reaction:   (Where dihydrogen is being produced.) So, at  and  pressure, the volume of gas that will be released when  of aluminium reacts. As we can see from the reaction equation that 2 moles of aluminium produce 3 moles of dihydrogen. i.e.,   reacts to give   At STP condition    and  ,  of Al gives  of . Therefore  of Al will give  of  i.e.,  of . At STP condition:  ...

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D Divya Prakash Singh
Given Pressure of 1g of an ideal gas A at  is . When 2g of another ideal gas B is introduced in the same flask at the same temperature, The pressure becomes . . We can assume the molecular masses of A and B be  respectively.  The ideal gas equation,  So, we have    and    Therefore,  or,  Hence the relation between the two gases is .      

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D Divya Prakash Singh
Given the condition : Temperature ,   for oxide gas and   for dinitrogen. As Density of gas is represented by,  And given that density is the same for both the gases at the same temperature condition. So, we have   (as R is constant} or,                       or,   
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