Home
Class 12
CHEMISTRY
The van der Waals equation for one mole ...

The van der Waals equation for one mole of a real gas can be written as `(P+(a)/(v^(2))(V-b)=RT`. For the gases `H_(2), NH_(3), and CH_(4)`, the value of 'a' `"bar L"^(-2)"mol"^(-2)` are 0.2453, 4.170 and 2.253` respectively.
Which of the following can be inferred from the 'a' values?

A

`NH_(3)` can be most easily liquified

B

`H_(2)` can be most easily liquified

C

value of 'a' for `CH_(4)` is less than that of `NH_(3)` because it has the lower molar mass

D

intermolecular forces are the strongest in hydrogen

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the van der Waals equation and the given values of 'a' for the gases H₂, NH₃, and CH₄. The van der Waals equation is given as: \[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \] Where: - \(P\) = pressure - \(V\) = volume - \(R\) = universal gas constant - \(T\) = temperature - \(a\) = measure of the attraction between particles - \(b\) = measure of the volume occupied by the gas particles ### Step 1: Understand the significance of 'a' The value of 'a' in the van der Waals equation represents the strength of the attractive forces between the gas molecules. A higher value of 'a' indicates stronger intermolecular attractions, which can lead to easier liquefaction of the gas. ### Step 2: Compare the values of 'a' for the given gases The values of 'a' for the gases are: - For H₂: \(a = 0.2453 \, \text{bar L}^{-2} \text{mol}^{-2}\) - For NH₃: \(a = 4.170 \, \text{bar L}^{-2} \text{mol}^{-2}\) - For CH₄: \(a = 2.253 \, \text{bar L}^{-2} \text{mol}^{-2}\) ### Step 3: Analyze the values - NH₃ has the highest value of 'a' (4.170), indicating it has the strongest intermolecular attractions. - CH₄ has a moderate value of 'a' (2.253), indicating weaker attractions than NH₃ but stronger than H₂. - H₂ has the lowest value of 'a' (0.2453), indicating it has the weakest intermolecular attractions. ### Step 4: Make inferences based on 'a' From the values of 'a', we can infer: 1. NH₃ can be most easily liquefied due to its high value of 'a'. 2. H₂ cannot be easily liquefied as it has the lowest value of 'a'. 3. The strength of intermolecular forces decreases in the order: NH₃ > CH₄ > H₂. ### Conclusion Based on the analysis, we conclude that the gas which can be most easily liquefied is NH₃, as it has the highest value of 'a'.
Promotional Banner

Similar Questions

Explore conceptually related problems

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

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

The van der Waal's equation of state for some gases can be expressed as : (P + (a)/( V^(2))) ( V - b) = RT Where P is the pressure , V is the molar volume , and T is the absolute temperature of the given sample of gas and a, b , and R are constants. Which of the following does not have the same dimensional formula as that for RT?

The Van der waal's equation for n moles of a real gas is given by (P+(n^(2)a)/V^(2)) (V-nb)=nRT , where P pressure of gas, V= volume of gas, T= temperature of gas R= molar gas constant, a & b= Van der waal's constant Which of the following have the same dimensions as those of nRT.

The van der Waals' constant 'a' for the gases O_(2), N_(2), NH_(3) "and " CH_(4) are 1.36, 1.39, 4.17 and 2.253 respectively, the gas which can be most easily liquefied is :

The equation of state for real gas is given by ((p + (a)/(V^(2))(V - b) = RT . The dimension of the constant a is ………………. .

The correct order of normal boiling of O_(2), N_(2), NH_(3) and CH_(4) for whom the values of van der Waals constant 'a' are 1.360, 1.390, 4.170 and 2.253 L^(2) atm mol^(-2) respectively, is:

The van der Waal equation of gas is (P + (n^(2)a)/(V^(2))) (V - nb) = nRT

The value of van der Waals constant a for the gases O_(2) , N_(2) , NH_(3) , and CH_(4) are 1.360 , 1.390 , 4.170 , and 2.253 L^(2) atm mol^(-2) , respectively. The gas which can most easily be liquefied is