3.1 Ideal gas and ideal gas equation of state

Two tanks contain methane. For the given conditions, methane can be treated as an ideal gas.

• Tank 1 has a volume of 0.3 m³, and is at a temperature of 20°C and a pressure of 300 kPa.
• Tank 2 contains 1.5 kg of methane, and is at a temperature of 30°C and a pressure of 800 kPa.

The partition between the two tanks is removed to allow methane in the tanks to mix and reach equilibrium. What is the equilibrium pressure if the temperature of the two tanks is 25°C at equilibrium?

Solution

Methane is treated as an ideal gas at the given conditions.

From Table G1: R=0.5182 kJ/kgK for methane.

Apply the ideal gas law $P\mathbb{V}=mRT$ to both initial and final conditions of methane in the two tanks.

Tank 1 at the initial condition:

$P_1\mathbb{V}_1=m_1RT_1$

$m_1 =\displaystyle\frac{P_1\mathbb{V}_1}{RT_1 }= \frac{300 \times 0.3}{0.5182 \times (273.15 + 20)}= 0.5925\ \rm{kg}$

Tank 2 at the initial condition:

$P_2\mathbb{V}_2=m_2RT_2$

$\mathbb{V}_2 =\displaystyle\frac{m_2RT_2}{ P_2}= \frac{1.5 \times 0.5182 \times (273.15 + 30)}{800}= 0.2945\ \rm{m^3}$

The two tanks are in equilibrium at the final state.

$m_3 = m_1 + m_2 = 0.5925 + 1.5 = 2.0925 \ \rm{kg}$

$\mathbb{V}_3 = \mathbb{V}_1 + \mathbb{V}_2 = 0.3 + 0.2945 = 0.5945 \ \rm{m^3}$

$P_3\mathbb{V}_3=m_3R_3T_3$

$P_3=\displaystyle\frac{m_3RT_3}{\mathbb{V}_3 }= \displaystyle\frac{2.0925 \times 0.5182 \times (273.15 + 25)}{0.5925}= 543.8 \ \rm{kPa}$

The equilibrium pressure of the two tanks at the final state is 543.8 kPa.

Important note:

• The temperature must be expressed in Kelvin when applying the ideal gas EOS.

Consider 1 kg of oxygen in a piston-cylinder device undergoing a thermodynamic cycle consisting of three processes.

• Process 1$\to$2: isochoric
• Process 2$\to$3: isothermal expansion
• Process 3$\to$1: isobaric compression

At state 1, T₁= 300 K, P₁=1.5 atm. At state 2, P₂= 3 atm. Treat oxygen as an ideal gas at the given conditions.

1. Sketch the cycle on a $P-v$ diagram.
2. Determine the temperature, T₂ , at state 2, and the specific volume, v₃, at state 3.

Solution

1. The cycle on a $P-v$ diagram

2. Oxygen is treated as an ideal gas at the given conditions.

From Table G1: R=0.2598 kJ/kgK for oxygen.

Apply the ideal gas law $Pv = RT$ to the three processes.

Process 1$\to$2 is an isochoric process; therefore, the specific volume remains constant in the process, $v_1=v_2$

$\because Pv = RT$

$\therefore v_1 = \displaystyle\frac{RT_1}{P_1}$ and  $v_2 = \displaystyle\frac{RT_2}{P_2}$

$\because v_1 = v_2$    and    $R = \rm{constant}$

$\therefore\displaystyle\frac{T_2}{T_1} = \displaystyle\frac{P_2}{P_1}$

$\therefore T_2 = T_1 \times \displaystyle\frac{P_2}{P_1} = 300 \times \frac{3}{1.5} = 600 \ \rm{K}$

Process 2$\to$3 is an isothermal expansion process; therefore, $T_3 = T_2 = 600 \ \rm{K}$.

Process 3$\to$1 is an isobaric compression process; therefore,

$P_3 = P_1 = 1.5 \ \rm{atm}=1.5 \times 101.325=152 \ \rm{kPa}$

$v_3 = \displaystyle\frac{RT_3}{P_3} = \displaystyle\frac{0.2598 \times 600}{152} = 1.026\ \rm{m^3/kg}$

The temperature at state 2 is 600 K and the specific volume at state 3 is 1.026 m³/kg.

Important note:

• The temperature must be expressed in Kelvin when applying the ideal gas EOS.