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Consider the van der Waals constants, a and b, for the following gases,

   Gas

Ar

Ne

Kr

Xe

a/(atm dm6 mol-2)

1.3

0.2

5.1

4.1

 b/(10-2 dm3 mol-1)

3.2

1.7

1.0

5.0


Which gas is expected to have the highest critical temperature?
1. Kr
2. Ne
3. Xe
4. Ar

1 Answer

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Best answer
Correct Answer - Option 1 : Kr

For every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature (Tc), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase.

Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures.

The formula for critical temperature is,

Critical temperature \( = \frac{{8a}}{{27Rb}}\)

Where, Tc = Critical Temperature,

a, b = van der Waals constant,

R = Universal Gas Constant

Here, in the given question, the van der Waals constant for each gas is given,

We know that the Universal Gas constant, R = 8.314 J/mol

For Ar; \({T_c} = \left( {\frac{8}{{27R}}} \right)\left( {\frac{a}{b}} \right)\)

\( = \left( {\frac{8}{{27R}}} \right)\left( {\frac{{1.3}}{{3.2}}} \right) = \left( {\frac{8}{{27R}}} \right)\left( {0.4} \right)\)

For Ne; \({T_c} = \left( {\frac{8}{{27R}}} \right)\left( {\frac{a}{b}} \right)\)

\( = \left( {\frac{8}{{27R}}} \right)\left( {\frac{{0.2}}{{1.7}}} \right) = \left( {\frac{8}{{27R}}} \right)\left( {0.12} \right)\)

For Kr; \({T_c} = \left( {\frac{8}{{27R}}} \right)\left( {\frac{a}{b}} \right)\)

\( = \left( {\frac{8}{{27R}}} \right)\left( {\frac{{5.1}}{{1.0}}} \right) = \left( {\frac{8}{{27R}}} \right)\left( {5.1} \right)\)

For Xe; \({T_c} = \left( {\frac{8}{{27R}}} \right)\left( {\frac{a}{b}} \right)\)

\( = \left( {\frac{8}{{27R}}} \right)\left( {\frac{{4.1}}{{5.0}}} \right) = \left( {\frac{8}{{27R}}} \right)\left( {0.82} \right)\)

Tc has order: Kr > Xe > Ar > Ne

There is no need to do the whole calculation as we are intending to find out the gas which has the greatest critical temperature. Since, the value of \(\left( {\frac{8}{{27R}}} \right)\) is constant in all calculations, the value multiplied by it should be considered. Here the value of  \(\frac{a}{b}\) is highest for Kr i.e., 5.1. Therefore, Kr has greatest value of critical temperature.

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