Chapter 4, Problem 75P

To determine

The volume of the second tank and the final equilibrium pressure.

Explanation of Solution

Given:

The pressure of the first tank $\left(P_1\right)$ is $350\text{kPa}$.

The  volume of the first tank $\left(V_1\right)$ is $1{\text{m}}^{\text{3}}$.

The temperature of the first tank $\left(T_1\right)$ is $10°\text{C}$.

The mass of the second tank $\left(m_2\right)$ is $3\text{kg}$.

The pressure of the second tank $\left(P_2\right)$ is $200\text{kPa}$.

The temperature of the second tank $\left(T_2\right)$ is $35°\text{C}$.

The final equilibrium temperature $\left(T\right)$ is $20°\text{C}$.

Calculation:

Refer Table A-1, “Molar mass, gas constant, and critical point properties”,

The gas constant of air $\left(R\right)$ as $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$.

Write the expression to obtain the mass of the first tank $\left(m_1\right)$.

  $m_1=\frac{P_1{V}_1}{R{T}_1}$

  $\begin{array}{c}{m}_1=\frac{\left(350\text{kPa}\right)\left(1{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(10°\text{C}\right)}\\ =\frac{\left(350\text{kPa}\right)\left(1{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(273+10\text{K}\right)}\\ =\frac{\left(350\text{kPa}\right)\left(1{\text{m}}^{\text{3}}\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(283\text{K}\right)}\\ =4.309\text{kg}\end{array}$

Write the expression to obtain the volume of the second tank $\left(V_2\right)$.

  $\begin{array}{c}{V}_2=\frac{m_{2}R{T}_2}{P_2}\\ V_2=\frac{\left(3\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(35°\text{C}\right)}{200\text{kPa}}\\ =\frac{\left(3\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(273+35\text{K}\right)}{200\text{kPa}}\\ =\frac{\left(3\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(308\text{K}\right)}{200\text{kPa}}\\ =1.326{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of the second tank is $1.326{\text{m}}^{\text{3}}$.

Write the expression to obtain the total volume $\left(V\right)$.

  $\begin{array}{c}V=V_1+V_2\\ V=1{\text{m}}^{\text{3}}+1.326{\text{m}}^{\text{3}}\\ =2.326{\text{m}}^{\text{3}}\end{array}$

Write the expression to obtain the total mass $\left(m\right)$.

  $\begin{array}{c}m=m_1+m_2\\ m=4.309\text{kg}+3\text{kg}\\ =7.309\text{kg}\end{array}$

Write the expression to obtain the final equilibrium pressure $\left(P\right)$.

  $P=\frac{mRT}{V}$

  $\begin{array}{c}P=\frac{\left(7.309\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(20°\text{C}\right)}{2.326{\text{m}}^{\text{3}}}\\ =\frac{\left(7.309\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(273+20\text{K}\right)}{2.326{\text{m}}^{\text{3}}}\\ =\frac{\left(7.309\text{kg}\right)\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(293\text{K}\right)}{2.326{\text{m}}^{\text{3}}}\\ =264\text{kPa}\end{array}$

Thus, the final equilibrium pressure is $264\text{kPa}$.