Question

2 g of hydrogen is sealed in a vessel of volume 0.02 m^3 and is maintained at 300 K. Calculate the pressure in the vessel.

Answer 1

PV = nRT
P = nRT/v
P= 1*0.08*300/0.02. { n = 2/2 }
P=1200 atm

Answer 2

2 g of hydrogen is sealed in a vessel of volume 0.02 m^3 and is maintained at 300 K. The pressure in the vessel is $1.24 \times 10^{5} \mathrm{Pa}$

Explanation:

Step 1:

Given data in the question

Hydrogen’s mass , m = 2 g

the vessel’s volume, $\mathrm{V}=0.02 \mathrm{m}^{2}$

vessel temperature , T = 300 K

the hydrogen molecular mass, M = 2 u

Numbers of moles, $n=\frac{m}{M}=\frac{2}{2}=1 \text { mole }$

Rydberg’s constant, $R=8.3 \mathrm{J} / \mathrm{Kmol}$

Step 2:

We get from the ideal gaseous equation

PV = nRT

$\begin{aligned}&P=\frac{n R T}{V}\\&P=\left(\frac{1 \times 8.3 \times 300}{0.02}\right)\\&P=\left(\frac{2490}{0.02}\right)\\&P=124500 \mathrm{Pa}\\&P=1.24 \times 10^{5} \mathrm{Pa}\end{aligned}$

Thus the pressure in the vessel is  $1.24 \times 10^{5} \mathrm{Pa}$.