The van der Waals equation for one mol of `CO_(2)` gas at low pressure will beA. `(P+(a)/(V^(2)))V=RT`B. `P(V-b)=RT-(a)/(V^(2))`C. `P=(RT)/(V-b)`D. `P=((RT)/(V-b)-(a)/(V^(2)))`

The van der Waals equation for one mol of `CO_(2)` gas at low pressure

1.	The van der Waals equation for one mol of `CO_(2)` gas at low pressure will beA. `(P+(a)/(V^(2)))V=RT`B. `P(V-b)=RT-(a)/(V^(2))`C. `P=(RT)/(V-b)`D. `P=((RT)/(V-b)-(a)/(V^(2)))`
Answer» `(P+(a)/(V^(2)))(V-b)=RT` At low pressure, volume is high and `b` may be ignored. So van der Waals equation becomes `(P+(a)/(V^(2)))V=RT` Hence, the answer is (`a`).

Discussion

No Comment Found

Related InterviewSolutions

The gases obey the different gas laws only theoretically. Practically all of them show some deviation from these laws. These are called real gases. The deviation are maximum under high pressyre and at low temperature. These are comparatively small when the conditions are reversed. It has been found that the easily liquefiable gases show more deviations from the ideal gas beheviour as compared to the gases which are liqufied with diffculty. Gas deviates from ideal gas beheviour because moleculesA. are coloulessB. attract each otherC. contain covalent bondD. show Brownian Movement.
The gases obey the different gas laws only theoretically. Practically all of them show some deviation from these laws. These are called real gases. The deviation are maximum under high pressyre and at low temperature. These are comparatively small when the conditions are reversed. It has been found that the easily liquefiable gases show more deviations from the ideal gas beheviour as compared to the gases which are liqufied with diffculty. An ideal gas is one which obeys the gas laws underA. a few selected experimental conditionsB. all experimental conditionsC. low pressure aloneD. high pressure alone.
Match the following gas laws with the equation representing them.
The density of mercury is `13.6 g mL^(-1)`. Calculate the approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.
The behaviour of matter in different state is governed by various physical law. According to you, what are the factors that determine the state of matter ?
When is deviation more in the behaviour of a gas from the ideal gas equation `PV=nRT`?A. At high temperature and low pressure.B. At low temperature and high pressure.C. At high temperature and high pressure.D. At low temperature and low pressure.
The behaviour of ideal gas is goverened by various gas laws which are described by mathematical statements as given below: (`i`) `PV=k` (constant) at constant `n` and `T` (`ii`) `V//T=k_(2)` (constant) at constant `n` and `P` (`iii`) `V//n=k_(3)` (constant) at constant `T` and `P` (`iv`) `PV=nRT` (`v`) `P//T=k_(4) (constant) at constant `n` and `V` Answer the following The value of `k_(2)` isA. Independent of nature and amount of gasB. Depends on temperature and pressure conditionsC. Depends on pressure and amount of gasD. Depends only on nature of gas
The behaviour of ideal gas is goverened by various gas laws which are described by mathematical statements as given below: (`i`) `PV=k` (constant) at constant `n` and `T` (`ii`) `V//T=k_(2)` (constant) at constant `n` and `P` (`iii`) `V//n=k_(3)` (constant) at constant `T` and `P` (`iv`) `PV=nRT` (`v`) `P//T=k_(4)` (constant) at constant `n` and `V` Answer the following A cylinder of `10 L` capacity at `300 K` containing the gas is used to fill balloons till finally the cylinder recorded a pressure of `10 m` bar. The number of `He` atoms still present in the cylinder isA. `4.82xx10^(21)`B. `2.41xx10^(23)`C. `2.41xx10^(21)`D. `4.82xx10^(23)`
Oxygen is present in a `1 L` flask at a pressure of `7.6xx10^(-10)mm Hg`. Calculate the number of oxygen molecules in the flask at `0^(@)C`.A. `2.7xx10^(9) "molecules"`B. `2.7xx10^(10) "molecules"`C. `2.7xx10^(11) "molecules"`D. `2.7 xx10^(12) "molecules"`
Consider the adjacent diagram. Initially, flask `A` contained oxygen gas at `27^(@)C` and `950 mm` of `Hg`, and flask `B` contained neon gas at `27^(@)C` and `900 mm`. Finally, two flask were joined by means of a narrow tube of negligible volume equipped with a stopcock and gases were allowed to mixup freely. The final pressure in the combined system was found to be `910 mm` of `Hg`. What is the correct relationship between volumes of the two flasks?A. `V_(B)=2V_(A)`B. `V_(B)=4V_(A)`C. `V_(B)=5V_(A)`D. `V_(B)=5.5V_(A)`

The van der Waals equation for one mol of `CO_(2)` gas at low pressure will beA. `(P+(a)/(V^(2)))V=RT`B. `P(V-b)=RT-(a)/(V^(2))`C. `P=(RT)/(V-b)`D. `P=((RT)/(V-b)-(a)/(V^(2)))`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment