Answer
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Hint: Compressibility factor is defined as the ratio of actual molar volume of gas to the calculated molar volume at the same pressure and temperature. It is denoted as Z.
Formula used: Compressibility factor is calculated as –
\[Z{\text{ }} = {\text{ }}\dfrac{{{V_{real}}}}{{{V_{ideal}}}}\]
where,
\[{{\text{V}}_{{\text{real}}}}\]= volume of real gas
\[{{\text{V}}_{{\text{ideal}}}}\]= volume of ideal gas
Complete step by step answer:
Real gases are referred to as non-ideal gases that occupy space and perform interactions. These gases do not obey the ideal gas equation, i.e., \[{\text{PV = nRT}}\]
Compressibility factor is defined as the ratio of actual molar volume of gas to the calculated molar volume at the same pressure and temperature. It is denoted as Z.
Compressibility factor is calculated as –
\[Z{\text{ }} = {\text{ }}\dfrac{{{V_{real}}}}{{{V_{ideal}}}}\]
where,
\[{{\text{V}}_{{\text{real}}}}\]= volume of real gas
\[{{\text{V}}_{{\text{ideal}}}}\]= volume of ideal gas
For an ideal gas, the compressibility factor, \[{\text{Z = 1}}\]
But as there are deviations from ideal gas behaviour or real gas behaviour. The compressibility factor or Z increases with pressure and decreases with temperature. When the pressure increases the compressibility factor is greater than one as \[{V_{real}}\] is greater than \[{V_{ideal}}\]
\[{{\text{H}}_{\text{2}}}\] and \[{\text{He}}\] are real gases and show real gas behaviour. They show more volume than usual.
Therefore, the compressibility factor of \[{H_2}\] and \[{\text{He}}\] is greater than one.
Hence, the correct answer is (B) i.e., \[{\text{ > 1}}\]
Note: A student can get confused between ideal gas and real gas. Ideal gas obeys all the gas laws under all conditions of temperature and pressure. On the other hand, a real gas is something that we observe around us. It does not obey the ideal gas laws all the time. But at a specific temperature known as Boyle’s Temperature, even a real gas behaves like an ideal gas.
Formula used: Compressibility factor is calculated as –
\[Z{\text{ }} = {\text{ }}\dfrac{{{V_{real}}}}{{{V_{ideal}}}}\]
where,
\[{{\text{V}}_{{\text{real}}}}\]= volume of real gas
\[{{\text{V}}_{{\text{ideal}}}}\]= volume of ideal gas
Complete step by step answer:
Real gases are referred to as non-ideal gases that occupy space and perform interactions. These gases do not obey the ideal gas equation, i.e., \[{\text{PV = nRT}}\]
Compressibility factor is defined as the ratio of actual molar volume of gas to the calculated molar volume at the same pressure and temperature. It is denoted as Z.
Compressibility factor is calculated as –
\[Z{\text{ }} = {\text{ }}\dfrac{{{V_{real}}}}{{{V_{ideal}}}}\]
where,
\[{{\text{V}}_{{\text{real}}}}\]= volume of real gas
\[{{\text{V}}_{{\text{ideal}}}}\]= volume of ideal gas
For an ideal gas, the compressibility factor, \[{\text{Z = 1}}\]
But as there are deviations from ideal gas behaviour or real gas behaviour. The compressibility factor or Z increases with pressure and decreases with temperature. When the pressure increases the compressibility factor is greater than one as \[{V_{real}}\] is greater than \[{V_{ideal}}\]
\[{{\text{H}}_{\text{2}}}\] and \[{\text{He}}\] are real gases and show real gas behaviour. They show more volume than usual.
Therefore, the compressibility factor of \[{H_2}\] and \[{\text{He}}\] is greater than one.
Hence, the correct answer is (B) i.e., \[{\text{ > 1}}\]
Note: A student can get confused between ideal gas and real gas. Ideal gas obeys all the gas laws under all conditions of temperature and pressure. On the other hand, a real gas is something that we observe around us. It does not obey the ideal gas laws all the time. But at a specific temperature known as Boyle’s Temperature, even a real gas behaves like an ideal gas.
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