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Question

The compressibility of a gas is less than unity at $$STP$$. Therefore,


A
Vm>22.4L
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B
Vm<22.4L
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C
Vm=22.4L
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D
Vm=44.8L
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Solution

The correct option is A $$V_m < 22.4 \:L$$
The compressibility factor '$$z$$' is the ratio of  the actual volume of the gas due to non ideal behaviour  to the volume of the gas due to ideal behaviour.
$$Z=\dfrac {V_{real}}{V_{ideal}}$$.
But $$Z<1$$.
Hence, $$\dfrac {V_{real}}{V_{ideal}}$$ or $$ {V_{real}} <{V_{ideal}}$$.
At $$STP$$, 1 mole of ideal gas occupies a volume of 22.4 $$L$$.
Hence, at $$STP$$, 1 mole of a real gas will occupy a volume less than 22.4 $$L$$.

Chemistry

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