Question

150 ml of dry hydrogen gas exarts 725 mmHg pressure at 25 degree celsius.calculate the volume of gas at NTP​

Answer 1

Answer:

131.09 mL

Explanation:

V₁ = 150 mL

P₁ = 725 mm Hg

T₁ = 25°C = (273+25)K = 298K

V₂ = ?

P₂ = 760 mm Hg (S.T.P.)

T₂ = 273 K (S.T.P.)

By Gas Equation, P₁V₁/T₁ = P₂V₂/T₂

=> 725 * 150 / 298 = 760 * V₂ / 273

=> 725 * 150 / 298 * 273 / 760 = V₂

Solve the equation...

V₂ = Volume at S.T.P. = 131.09 mL

Answer 2

Given:

Volume of dry hydrogen gas $\[{V_1} = 150\,ml\]$

Pressure $\[{P_1} = 725\,mm\,Hg\]$

Temperature $\[{T_1} = 25^\circ C = 298.15K\]$

To Find: Volume of gas at NTP

Solution:

At NTP, the temperature and pressure are taken to be $\[293.15{\text{ }}K\;\,\]$ and $\[1\,atm\]$.

Therefore,

$\[{T_2} = 293.15{\text{ }}K\;\,\]$

$\[{P_2} = 1\,atm = 760\,mm\,Hg\]$

According to the gas equation,

$\[\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\]$

$\[\frac{{725\,mm\,Hg \times 150\,ml}}{{298.15K}} = \frac{{760\,mm\,Hg \times {V_2}}}{{293.15{\text{ }}K\;\,}}\]$

Therefore, the volume of gas at NTP​ can be given as:

$\[{V_2} = \frac{{725\,mm\,Hg \times 150\,ml \times 293.15{\text{ }}K}}{{298.15K \times 760\,mm\,Hg}}\]$

$\[{V_2} = 140.69 \approx 140.7\,ml\]$

Hence, the volume of gas at NTP is $\[140.7\,ml\]$.