Question

A welder using a tank of volume$\mathbf{0}\mathbf{.}\mathbf{0750}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$fills it with oxygen (molar mass 32.0 g/mol) at a gauge pressure of$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{*}{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$and temperature of$\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is$\mathbf{22}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$, the gauge pressure of the oxygen in the tank is$\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{*}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$. Find (a) the initial mass of oxygen and (b) the mass of oxygen that has leaked out.

Short answer

(a)The initial mass of oxygen before the leak is 0.37 kg .

(b)The mass of oxygen leaked out is 0.1kg.

Step-by-step solution

The expression for initial and leaked mass of oxygen

The number of moles is $n_i=\frac{m_i}{M}$, from which the initial mass is $m_i=n_{i}M{o}_2$.

Molecular mass of oxygen is $M_{o_2}=32.0g/mol$.

The number of moles$n_i$ is found from the ideal gas equation. Expressing the ideal gas equation in terms of absolute initial pressure, $P_{i}V=n_{i}R{T}_i$. Rearranging to find initial number of moles as, $n_i=\frac{P_{i}V}{R{T}_i}$in which the initial pressure is$P_i=P_{abc}+P_{g,i}$ where$P_{abc}$ is the absolute pressure and $P_{g,j}$, initial gauge pressure. Then,

$n_i=\frac{(P_{abc}+P_{g,j})*V}{R{T}_i}$

Similarly, the final number of moles become$n_f=\frac{P_{f}V}{R{T}_f}$ in which the final pressure expressed as $P_f=P_{abc}+P_{g,j}$, where$P_f$ is,$P_{abc}$ is the absolute pressure,$P_{g,f}$ is the final gauge pressure. Then,

$n_f=\frac{(P_{abc}+P_{g,f})V}{R{T}_f}$

Then the final mass of oxygen becomes $m_f=n_f*M_O$ .

The mass of leaked oxygen is then the difference between initial and final value of masses.

$\mathrm{Dm}=m_i-m_f$

Calculating the initial mass of oxygen.

(a)The initial mass is $m_t=n_t{M}_{O_2}$.

$M_{O_2}=32.0g/mol$.

Substitute the values of role="math" localid="1668233742784" $P_{abs}=1*{10}^5\mathrm{Pa},P_{g,i}=3*{10}^5\mathrm{Pa},V=0.075{\mathrm{m}}^3$$R=8.314\mathrm{J}/\text{mol.}K,T_i={37}^{\circ }\mathrm{C}=37+273=310\mathrm{K}\text{in}{n}_i=\frac{\left(P_{cbs}+P_{g,i}\right)\cdot V}{R{T}_i}$,

$\begin{array}{l}{n}_i=\frac{\left[\right(1*{10}^5)+(3*{10}^5\left)\right]*0.075}{8.314+310}\\ n_i=11.64moles\end{array}$

Then, the initial mass is obtained by substituting$M_{O_2}=32.0g/mol$ and$n_i=11.64moles$ in $m_i=n_i{M}_{O2}$

$$\begin{gathered} m_i=11.64*32 \\ {m}_i=372.5g \\ {m}_i=0.37kg \end{gathered}$$

Thus, the initial mass of oxygen before leaking process is 0.37kg.

Step 3:Calculating the final mass of oxygen

The final mass of oxygen is $m_f=n_f*M_{O2}$.

$M_{O2}=32.0g/mol$.

Find the final number of moles by substituting the values of $P_{abc}=1*{10}^{5}Pa$, $P_{g.f}=3^{\ast }{10}^5\mathrm{PaV}=0.075{\mathrm{m}}^3,R=8.314\mathrm{J}/\mathrm{mol}.\mathrm{K},T_f={22.0}^{\circ }\mathrm{C}=22+273=295\mathrm{K}$ in

$n_f=\frac{\left(P_{abc}+P_{g.f}\right)V}{R{T}_f}$

$$\begin{gathered} n_f=\frac{\left[\right(1*{10}^5)+(1.8*{10}^5\left)\right]*0.075]}{8.314*295} \\ {n}_f=8.56mol \end{gathered}$$

Find the final mass by substituting the above values for$n_f=8.56mol$ and$M_{O_2}=32.0g/mol$ in $m_f=n_f*M_{o_2}$.

$$\begin{gathered} m_f=8.56*32.0 \\ {m}_f=273.92g \\ {m}_f=0.27kg \end{gathered}$$

Thus, the final mass of oxygen after the leak is 0.27 kg.

Calculating the leaked mass

(b)The mass of leaked oxygen is $\mathrm{Dm}=m_i-m_f$. Substitute the value ofdata-custom-editor="chemistry" $m_{\mathrm{i}}=0.37kg$ and$m_f=0.27kg$

$$\begin{gathered} Dm=0.37-0.27 \\ Dm=0.10kg \end{gathered}$$

Thus, the mass of oxygen leaked out is 0.10 kg.