Question

A rigid tank of volume \(0.75 \mathrm{~m}^{3}\) is initially evacuated. \(\mathrm{A}\) hole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) flows in until the pressure in the tank reaches \(1 \mathrm{bar}\). Heat transfer between the contents of the tank and the surroundings is negligible. Determine the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\).

Short answer

Final temperature in the tank is \( 25^{\circ}\text{C}\).

Step-by-step solution

- Understand the Problem

The problem involves calculating the final temperature inside a rigid tank after it has been filled with air from the surroundings until the pressure equilibrates at 1 bar. Initially, the tank is evacuated, meaning it contains no air.

- Note Known Variables

Given data: - Volume of the tank: \(V = 0.75 \,\mathrm{m}^3\)- Initial pressure in the tank: \(P_{1} = 0 \,\mathrm{bar}\)- Final pressure in the tank: \(P_{2} = 1 \,\mathrm{bar}\)- Surrounding pressure: \(1 \, \mathrm{bar}\)- Surrounding temperature: \( 25^{\circ}C\) or \( T_{2} = 25 + 273.15 = 298.15 \,\mathrm{K} \)

- Apply the Ideal Gas Law

Since the heat transfer is negligible and assuming the gas behaves ideally, the air that flows into the tank will reach thermal equilibrium with the surrounding air. The Ideal Gas Law is \[ PV = nRT \]Where:- \( P\) = pressure- \( V\) = volume- \( n\) = number of moles - \( R\) = universal gas constant (8.314 J/(mol·K))- \( T\) = temperatureFor the initial evacuated state: \( PV = 0 \), thus we ignore initial conditions for gas.

- Equate Final Conditions

Let's assume the final state as equilibrium:For final conditions, \[ P_{2} V = n_{2} R T_{2} \]Where:\( P_{2} = 1 \, \mathrm{bar} = 100,000 \,\mathrm{Pa}\), \( V = 0.75 \, \mathrm{m}^{3}\), and \( R = 8.314 \,\mathrm{J/(mol·K)}\).

- Calculate the Final Temperature

Given that the pressure \( P_{2} = 100,000 \,\mathrm{Pa} \) remains constant and the tank is rigid, the final temperature is that of air flowing in from surroundings. Thus \[ T_{2} = 298.15 \, \mathrm{K} = 25^{\circ}C \]

Key concepts

rigid tank

A rigid tank is a container that does not change its volume, regardless of the pressure or temperature inside it. In our exercise, the volume of the tank is fixed at 0.75 m³. This means that during the process of air filling the tank, the volume remains constant.

thermal equilibrium

Thermal equilibrium is a state when two systems or objects reach the same temperature and thus no heat flows between them. In the given exercise, when air flows into the tank from the surroundings, it eventually reaches thermal equilibrium with the surrounding environment. The final temperature of the air in the tank will be the same as the surrounding temperature, which is given as 25°C.

ideal gas behavior

Ideal gas behavior is an assumption where the gas behaves according to the Ideal Gas Law, which is \( PV = nRT \). This law relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T). Here, we assumed that the air behaves ideally, meaning the gas particles do not interact and the volume of the particles themselves is negligible.

pressure equilibrium

Pressure equilibrium means that the pressure inside the tank equalizes with the pressure of the surroundings. Initially, the tank is evacuated with a pressure of 0 bar. A hole allows air to enter until the pressure inside reaches 1 bar, the same as the outside pressure. This is the pressure equilibrium state.