Question

A steel tank contains$\mathbf{300}\mathbf{}\mathbf{g}\mathbf{}$ of ammonia gas (${\mathbf{NH}}_3$ ) at a pressure of$\mathbf{1}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^6\mathbf{pa}$ and a temperature of $\mathbf{77}°\mathbf{C}$.

(a) What is the volume of the tank in liters?

(b) Later the temperature is $\mathbf{22}\mathbf{°}\mathbf{C}$and the pressure is$\mathbf{8}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^5\mathbf{}\mathbf{pa}$ . How many grams of gas have leaked out of the tank?

Short answer

1. The tank has a volume of $38L$.
2. The amount of gas that leaked out of the tank is$71\text{g}$.

Step-by-step solution

Concept

The ideal gas law states thatfor the change from the initial to the final state, the product of pressure, volume and reciprocal of absolute temperature, remains constant.It is given as-

$\frac{\mathrm{pV}}{T}\mathbf{=}\mathbf{constant}$

The ideal gas equation is given as-

$\mathbf{pV}\mathbf{=}\mathbf{nRT}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(\mathbf{1}\right)$

Here,Pis pressure, V is volume,Tis temperature, R is the gas constant and is number of moles.

Step 2: Given Data

1. The mass of ammonia $M_s=300\text{gm}$
2. The initial pressure of the gas$p_1=1.35\times {10}^6\text{\hspace{0.17em}Pa}$
3. The initial temperature of the gas$T_1=77°\text{C}=350\text{\hspace{0.17em}K}$
4. The final pressure of the gas$p_2=8.7\times {10}^5\text{\hspace{0.17em}Pa}$
5. The final temperature of the gas $T_2=22\text{\hspace{0.17em}°C}=295\text{\hspace{0.17em}K}$
6. The molar mass of ammonia $M=17\text{g}/\text{mol}$

Calculations

a.

The number of moles of ammonia gas are:

$n_1=\frac{\text{samplemass}\left(M_s\right)}{\text{molarmass}(M)}$

$\begin{array}{c}{n}_1=\frac{\text{300\hspace{0.17em}g}}{\text{17\hspace{0.17em}g}/\text{mol}}\\ =17.65\text{\hspace{0.17em}mol}\end{array}$

Ammonia gas obeys the ideal gas equation.

$\begin{array}{l}pV=nRT\\ V_1=\frac{n_{1}R{T}_1}{P_1}\end{array}$

For the given values,

$\begin{array}{c}{V}_1=\frac{17.65\text{\hspace{0.17em}mol}\times 8.31\text{\hspace{0.17em}J}/\text{K}\cdot \text{mol}\times 350\text{\hspace{0.17em}K}}{1.35\times {10}^6\text{\hspace{0.17em}Pa}}\\ =0.038{\text{\hspace{0.17em}m}}^3\\ =38\text{\hspace{0.17em}L}\end{array}$

The tank’s volume is $38\text{L}.$.

b.

Let the volume of the gas in final state is $V_2$. Thenumber of moles of the gas inside the tank is-

$n_2=\frac{p_2{V}_1}{R{T}_2}$

For the given values,

$\begin{array}{c}{n}_2=\frac{8.7\times {10}^5\text{\hspace{0.17em}Pa}\times 0.038{\text{\hspace{0.17em}m}}^3}{8.31\text{\hspace{0.17em}J}/\text{K}\cdot \text{mol}\times 295\text{\hspace{0.17em}K}}\\ =13.45\text{\hspace{0.17em}mol}\end{array}$

Thus, the change in number of moles of the gas is

$\begin{array}{c}\Delta n=n_1-n_2\\ =17.65\text{\hspace{0.17em}mol}-13.45\text{\hspace{0.17em}mol}\\ =\text{4.2\hspace{0.17em}mol}\end{array}$

Next, we determine the change in the mass of the gas as

$\Delta M_s=\Delta n\times M$

$\begin{array}{c}\Delta {\text{M}}_s=4.2\text{\hspace{0.17em}mol}\times 17\text{\hspace{0.17em}g}/\text{mol}\\ =71.4\text{\hspace{0.17em}g}\end{array}$

Thus, the amount of gas leaked is $71.4\text{\hspace{0.17em}gm}$

Step 4: Conclusion

The volume of the tank of ammonia is $38\text{L}$and the amount gas that leaked from the tank is $71.4\text{\hspace{0.17em}gm}$.