Three gases, methane, ammonia and water vapour which have comparable molecular masses are liquefied at the same temperature. However,the pressure required to be applied is different for the three gases. Justify.

Open in App

Solution

The greater the intermolecular force of attraction, the lower the pressure required to liquefy a gas.
In $H_{2} O$ and $N H_{3},$ there exists a strong hydrogen bonding, and hence, the intermolecular forces of attraction are stronger.
Between $N H_{3}$ and $H_{2} O,$ water has stronger intermolecular forces of attraction due to stronger hydrogen bonding with oxygen (more electronegative).
$C H_{4}$ being a non-polar molecule has only Van der Waal's forces of attraction, which are weak forces.
Hence, the order of intermolecular forces of attraction is $C H_{4} < N H_{3} < H_{2} O$ and the pressure required to be applied increases in the order: $H_{2} O < N H_{3} < C H_{4} .$

Suggest Corrections

Similar questions

Why gases such as hydrogen and heleium are not liquefied at room temperature by applying very high pressure?

O_{2}

\begin{matrix} Gas & C. Temp (K) & C. Pressure (atm) P & 5.1 & 2.2 Q & 33 & 13 R & 126 & 34 S & 135 & 40 \end{matrix}

1869

C O_{2}

(T_{c})

(P_{c)}

V_{c} = 3 b, P_{c} = \frac{a}{27 b^{2}}, T_{c} = \frac{8 a}{27 R b}

^{'} b^{'}

If gases liquifies by applying high pressure and low temperature then why water vapours turn into water without applying any pressure?

Join BYJU'S Learning Program