
The compressibility factor for \[{H_2}\] and He is usually:
A. > 1
B. = 1
C. < 1
D. either of these
Answer
569.1k+ views
Hint: We know that the other name of compressibility factor is compression factor or gas deviation factor. Compressibility factor is defined as the ratio of the actual volume at a given temperature to the ideal volume under the same conditions of pressure and temperature.
Complete answer:
Since there are no attractive or repulsive forces between the molecules of an ideal gas, the actual volume will be the same as ideal volume. Hence, the compressibility factor is equal to 1.
In case of real gases, forces between the molecules are a function of the distance between them. When the molecules are far apart, there are attractive forces between them but when they are forced together their electronic fields overlap due to which repulsive forces come into play.
Pressure and molecule attraction tend to confine the molecules while temperature and molecular repulsion tend to separate them. When the temperature of the gas is high, the kinetic energy of the molecules tends to move apart. So at high temperatures, the actual volume of the gas will be more than the ideal volume. Hence, the compressibility factor will be greater than one at high temperatures.
The compressibility factor (Z) for real gases is given by the formula shown below.
\[Z = 1 + \dfrac{{Pb}}{{RT}} - \dfrac{a}{{VRT}}\]
At high temperature, T = temperature is high due to which value of \[\dfrac{a}{{VRT}}\] decreases. Also at high temperature, pressure is high due to which the value of \[\dfrac{{Pb}}{{Rt}}\] increases in first due to which the net value of Z becomes greater than 1.
Therefore, we conclude that for \[{H_2}\] and He, the compressibility factor is greater than one.
Hence, the correct option is (A).
Note: Compressibility factor is very useful for modifying the ideal gas law to account for the real gas behavior. One can easily get confused between real gas and ideal gas. As we know that the ideal gas follows all the gas laws at all conditions of pressure and temperature, whereas real gas does not.
Complete answer:
Since there are no attractive or repulsive forces between the molecules of an ideal gas, the actual volume will be the same as ideal volume. Hence, the compressibility factor is equal to 1.
In case of real gases, forces between the molecules are a function of the distance between them. When the molecules are far apart, there are attractive forces between them but when they are forced together their electronic fields overlap due to which repulsive forces come into play.
Pressure and molecule attraction tend to confine the molecules while temperature and molecular repulsion tend to separate them. When the temperature of the gas is high, the kinetic energy of the molecules tends to move apart. So at high temperatures, the actual volume of the gas will be more than the ideal volume. Hence, the compressibility factor will be greater than one at high temperatures.
The compressibility factor (Z) for real gases is given by the formula shown below.
\[Z = 1 + \dfrac{{Pb}}{{RT}} - \dfrac{a}{{VRT}}\]
At high temperature, T = temperature is high due to which value of \[\dfrac{a}{{VRT}}\] decreases. Also at high temperature, pressure is high due to which the value of \[\dfrac{{Pb}}{{Rt}}\] increases in first due to which the net value of Z becomes greater than 1.
Therefore, we conclude that for \[{H_2}\] and He, the compressibility factor is greater than one.
Hence, the correct option is (A).
Note: Compressibility factor is very useful for modifying the ideal gas law to account for the real gas behavior. One can easily get confused between real gas and ideal gas. As we know that the ideal gas follows all the gas laws at all conditions of pressure and temperature, whereas real gas does not.
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