You visited us 1 times! Enjoying our articles? Unlock Full Access!

Compressibility Factor

If the compre...

Question

If the compressibility factor of $H_{2}$ is $1.8$ , then the volume of one mole of real $H_{2}$ at STP will be:

40.32 L

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

22.42 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30.322 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12.44 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $40.32 L$
Compressibility factor is used as a correction factor to ideal behaviour.
$V_{r e a l} = Z V_{i d e a l}$ (Z = compressibility factor)
$V_{r e a l} = 1.8 \times 22.4 L$
$V_{r e a l} = 40.322 L$

Suggest Corrections

Similar questions

H_{2}

1.8

H_{2}

H_{2}

Z = \frac{P V}{R T}

1 mole

1 atm

273.15 K

H_{2}

100

H_{2}

Join BYJU'S Learning Program