91Ó°ÊÓ

Ideal gas law states the mathematical relation or equation of a state characterized by macroscopic quantities pressure, temperature and volume of a hypothetical ideal gas.

Mathematically,

PV = nRT ….. (1)

Here,is the pressure,is the volume,is the number of moles,is the universal gas constant, andis the temperature.

Consider the given data as below.

The height, h = 1.00 m

The diameter, d = 0.120 m

The radius, $\mathrm{r}=\frac{\mathrm{d}}{2}=\frac{0.120}{2}=0.06\mathrm{m}$

Molar mass, M = 44.1 g/mol =$44.1×{10}^{-3}\mathrm{kg}/\mathrm{mol}$

The universal gas constant, $\mathrm{R}=8.314\mathrm{J}/\mathrm{mol}.\mathrm{k}$

Temperature, $\mathrm{T}=22°\mathrm{C}=\left(22+273.15\right)\mathrm{K}=297.15\mathrm{K}$

Initial the gauge pressure is $1.30×{10}^{6}Pa$

Final gauge pressure,$3.40××{10}^5\mathrm{Pa}$

From equation (1), the number of moles can be written as,

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

Where is the mass of gas given and is the atomic weight.

So, the equation of state is,

role="math" localid="1668182854367" $\begin{array}{r}\mathrm{PV}=\left(\frac{\mathrm{m}}{\mathrm{M}}\right)\mathrm{RT}\left(2\right)\\ \frac{\mathrm{P}}{\mathrm{m}}=\frac{\mathrm{RT}}{\mathrm{MV}}\\ \frac{\mathrm{P}}{\mathrm{m}}=\mathrm{constant}\\ \mathrm{Therefore},\\ \frac{{\mathrm{p}}_1}{{\mathrm{m}}_1}=\frac{{\mathrm{p}}_2}{{\mathrm{m}}_2}\left(3\right)\\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{tank}\mathrm{is},\\ \mathrm{V}=\mathrm{hA}\\ ={\mathrm{³óÏ€°ù}}^2\\ =(1.00\mathrm{m})×3.14×(0.060\mathrm{m}{)}^2\\ =0.01131{\mathrm{m}}^3\end{array}$

Solve for m₁ from equation (2),

$\begin{array}{r}{\mathrm{m}}_1=\frac{{\mathrm{P}}_1\mathrm{VM}}{\mathrm{RT}}\\ \mathrm{Substitute}\mathrm{known}\mathrm{values},\mathrm{and}\mathrm{you}\mathrm{have}\\ {\mathrm{m}}_1=\frac{\left(1.40×{10}^6\right)(0.01131)\left(44.1×{10}^{−3}\right)}{8.314×(22.0+273.15)}\\ =0.2845\mathrm{kg}\\ \mathrm{Now}\mathrm{rearrange}\mathrm{equation}\left(3\right)\mathrm{as}\mathrm{below}.\\ {\mathrm{m}}_2={\mathrm{m}}_1\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)\end{array}$

Substitute known values in the above equation.

$\begin{array}{r}{\mathrm{m}}_2=\left(0.2845\mathrm{kg}\right)\left(\frac{4.41×{10}^5\mathrm{Pa}}{1.40×{10}^6\mathrm{Pa}}\right)\\ =0.0896\mathrm{kg}\end{array}$

So,${\mathrm{m}}_2$ is the mass left in the tank.

Therefore. the mass of propane used up is,

$\begin{array}{r}{\mathrm{m}}_1−{\mathrm{m}}_2=0.2845\mathrm{kg}−0.0896\mathrm{kg}\\ =0.195\mathrm{kg}\end{array}$

Hence, the mass of propane used up is 0.195 kg.