Question

Volume of one gram equivalent of ${H_2}$ at NTP is:
(a)5.6L
(b)11.2L
(c)22.4L
(d)44.8L

Answer 1

Hint: Volume is the third way to measure the amount of matter, after item count and mass. Volume varies greatly depending on the density of the substances. Avogadro explained that the volumes of all gases can be easily determined.

Complete answer:
Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal number of particles. Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another. This decreases the amount of empty spaces and reduces the volume of gas. Gas volume is also affected by temperature. When a gas is heated, its molecules move faster and the gas expands. Because of the variation in gas volume due to pressure and temperature changes, the comparison is done at STP (Standard temperature and pressure) and Normal temperature and pressure (NTP).
The volume of $1$ mole of ${H_2}$ at NTP is = $22.4L$
For, $1g$ of ${H_2}$, the volume at NTP= $\dfrac{{22.4}}{2} = 11.2L$
In many books the volume of gas in STP is equal to the volume of gas at NTP.
And hence the correct answer is option b.

Note:
Do not consider the temperature as $273K$, while solving the ideal gas equation. As the temperature taken in NTP is $293K$, whereas for STP it is $273K$.
The condition for STP is:
Pressure= $1$ atm and temperature= $273K$
According to ideal gas equation, $PV = nRT$
$V = \dfrac{{273K*0.08205{L_{}}atm/(mol.K)}}{{1atm \times 1mol}}$
$V = 22.4L$
For NTP, the conditions are:
Pressure= $1$ atm and temperature= $293K$
$V = \dfrac{{293K*0.08205{L_{}}atm/(mol.K)}}{{1atm \times 1mol}}$
$V = 24.02L$