Question

A \(1-m^{3}\) tank containing air at \(10^{\circ} \mathrm{C}\) and 350 kPa is connected through a valve to another tank containing \(3 \mathrm{kg}\) of air at \(35^{\circ} \mathrm{C}\) and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at \(20^{\circ} \mathrm{C}\). Determine the volume of the second tank and the final equilibrium pressure of air. Answers: \(1.33 \mathrm{m}^{3}, 264 \mathrm{kPa}.\)

Short answer

Answer: The volume of the second tank is approximately 1.33 m³, and the final equilibrium pressure is approximately 264 kPa.

Step-by-step solution

Gather known data.

For tank 1, we are given its volume, temperature, and pressure: - \(V_1 = 1\,\text{m}^3\) - \(T_1 = 10^\circ\,\text{C} = 283.15\,\text{K}\) (converted to Kelvin) - \(P_1 = 350\,\text{kPa}\) For tank 2, we are given its mass of air, temperature, and pressure: - \(m_2 = 3\,\text{kg}\) - \(T_2 = 35^\circ\,\text{C} = 308.15\,\text{K}\) (converted to Kelvin) - \(P_2 = 200\,\text{kPa}\) The final temperature after thermal equilibrium is reached will be the same as the surrounding temperature: - \(T_f = 20^\circ\,\text{C} = 293.15\,\text{K}\) (converted to Kelvin) Moving forward, we want to find the final equilibrium pressure and the volume of the second tank.

Apply ideal gas law to each tank.

To calculate the gas constant (R) for air, divide the universal gas constant (\(8314 \,\text{J}/\text{kmol}\cdot\text{K}\)) by the molar mass of air (29.92 kg/kmol), yielding \(R = 277.483 \,\text{J}/\text{kg}\cdot\text{K}\). For tank 1, apply the ideal gas law, \(P_1V_1 = m_1RT_1\) We can rearrange the formula to calculate the mass of air in tank 1: \(m_1 = \frac{P_1V_1}{RT_1} = \frac{(350\,\text{kPa})(1\,\text{m}^3)}{(277.483\,\text{J}/\text{kg}\cdot\text{K})(283.15\,\text{K})} = 4.432\,\text{kg}\) For tank 2, we also apply the ideal gas law: \(P_2V_2 = m_2RT_2\) However, we haven't found the volume of the second tank yet, so we'll return to this equation later.

Apply mass conservation principle.

Since no mass is added or removed from the combined system, we can write the mass conservation equation: \(m_f = m_1 + m_2 = 4.432\,\text{kg} + 3\,\text{kg} = 7.432\,\text{kg}\)

Calculate final volume and pressure.

The final volume of the combined system \(V_f = V_1 + V_2\). Now, apply the ideal gas law to the combined system: \(P_fV_f = m_fRT_f\) We can rearrange the equation to solve for \(P_f\): \(P_f = \frac{m_fRT_f}{V_f} = \frac{(7.432\,\text{kg})(277.483\,\text{J}/\text{kg}\cdot\text{K})(293.15\,\text{K})}{V_f}\) Returning to the ideal gas law for tank 2, we can substitute the known values for \(P_f\) and \(T_f\) (thermal equilibrium): \(P_2V_2 = m_2RT_f\) Rearranging to solve for \(V_2\): \(V_2 = \frac{m_2RT_f}{P_2} = \frac{(3\,\text{kg})(277.483\,\text{J}/\text{kg}\cdot\text{K})(293.15\,\text{K})}{200\,\text{kPa}} = 1.329\,\text{m}^3\) Next, we substitute back into our equation for \(P_f\): \(P_f = \frac{(7.432\,\text{kg})(277.483\,\text{J}/\text{kg}\cdot\text{K})(293.15\,\text{K})}{1\,\text{m}^3 + 1.329\,\text{m}^3} = 263.83\,\text{kPa}\) (rounded to 264 kPa) The volume of the second tank is \(V_2 = 1.33\,\text{m}^3\) and the final equilibrium pressure is \(P_f = 264\,\text{kPa}\).

Key concepts

Thermodynamics and the Ideal Gas Law

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It encompasses a variety of laws and principles that describe how energy is transferred within a system and between systems. One of the fundamental equations in thermodynamics is the ideal gas law, given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvins.

The equation assumes that the gas being described behaves ideally, meaning that the gas particles are small compared to the space between them, are in constant, random motion, and have perfectly elastic collisions with the walls of their container. In practical applications, the ideal gas law provides valuable information about how changes in pressure, volume, and temperature affect a gas. Using this equation, we can predict one variable if the others are known, making it essential for understanding the behavior of gases in various scientific and engineering applications.

Thermal Equilibrium

When two systems are in contact and can exchange heat or work, they will eventually reach a state where their temperatures are equal. This condition is known as thermal equilibrium, a concept crucial for understanding how energy is distributed in a thermodynamic process.

At thermal equilibrium, entropy is maximized for an isolated system, meaning that the energy within the system is distributed as evenly as possible. Once reached, there is no net flow of thermal energy between the bodies in question. In the context of the ideal gas law, reaching thermal equilibrium will involve adjusting the pressure and volume of gases involved. According to the zeroth law of thermodynamics, if two systems are in thermal equilibrium with a third system separately, they are in thermal equilibrium with each other. This principle enables us to use thermometers as the intermediate system to measure temperature and infer equilibrium conditions.

Mass Conservation Principle

The principle of mass conservation, also known as the law of conservation of mass, states that mass cannot be created or destroyed in an isolated system through physical changes or chemical reactions. In other words, the mass of the system must remain constant over time as long as the system is closed to all transfers of matter.

When applying this principle to thermodynamic processes involving ideal gases, as in the exercise above, the total mass of gas in all containers must remain the same before and after they reach thermal equilibrium, provided no gas is added or lost. This fundamental concept allows us to predict the behavior of the total mass of gases when they are allowed to mix — enabling us to solve for unknown quantities like volume and pressure in combined systems by ensuring that the mass before and after the process is the same.