Question

A rigid tank contains \(2 \mathrm{kg}\) of an ideal gas at \(4 \mathrm{atm}\) and \(40^{\circ} \mathrm{C}\). Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is \(2.2 \mathrm{atm},\) the final temperature in the \(\operatorname{tank}\) is \((a) 71^{\circ} \mathrm{C}\) \((b) 44^{\circ} \mathrm{C}\) \((c)-100^{\circ} \mathrm{C}\) \((d) 20^{\circ} \mathrm{C}\) \((e) 172^{\circ} \mathrm{C}\)

Short answer

Question: An ideal gas is initially stored in a rigid tank with a mass of 2 kg, a volume of V, and a temperature of 40°C. When the valve is opened, half of the gas is allowed to escape, and the final pressure is 2.2 atm. Using the ideal gas law, find the final temperature of the remaining gas. Answer: To find the final temperature, the solution requires three main steps: 1) calculating the initial moles and volume, 2) calculating the final moles after half the mass has escaped, and 3) using the ideal gas law to find the final temperature after the escape of gas.

Step-by-step solution

Calculate Initial Moles and Volume

First we use the ideal gas law: PV = nRT. We are given the initial pressure P1=4 atm, mass m1=2 kg and initial temperature T1=40C. Also given is the Gas Constant, R = 8.314 J/(K·mol). First, let's find the initial moles n1. To do that, we need to divide the initial mass m1 = 2 kg by the molar mass Mi of the ideal gas. Assuming air to be the ideal gas, the molar mass Mi = 28.97 g/mol or 0.02897 kg/mol. This gives us n1. $$ n_{1} = \frac{m_{1}}{M_i} = \frac{2 \, \mathrm{kg}}{0.02897 \, \mathrm{kg/mol}} $$ Then, using the ideal gas law, we can find the initial volume V of the gas. $$ V = \frac{n_{1}RT_{1}}{P_{1}} = \frac{n_{1} \times 8.314 \, \mathrm{J/(K·mol)} \times (40+273.15) \, \mathrm{K}}{4\, \mathrm{atm}} $$ Since 1 atm = 101.325 kPa, $$ V = \frac{n_{1}RT_{1}}{P_{1}} = \frac{n_{1} \times 8.314 \, \mathrm{J/(K·mol)} \times (40+273.15) \, \mathrm{K}}{4 \times 101.325\, \mathrm{kPa}} $$

Calculate Final Moles

We are told that half of the mass of the gas escapes the tank. Therefore, the final mass m2 = m1/2. Since the molar mass Mi stays the same, we can find the final moles n2. $$ n_{2} = \frac{m_{2}}{M_i} = \frac{m_{1}}{2M_i} $$

Calculate the Final Temperature

We now have all the variables we need to calculate the final temperature T2. We know the final pressure P2 = 2.2 atm, and we can use the ideal gas law once more, adjusting the pressure unit. $$ T_{2} = \frac{P_{2}V}{n_{2}R} = \frac{2.2 \times 101.325\, \mathrm{kPa} \times V}{n_{2} \times 8.314 \, \mathrm{J/(K·mol)}} $$ We calculate T2 in Kelvin and convert it into Celsius. $$ T_{2} (\mathrm{Celsius}) = T_{2} (\mathrm{Kelvin}) - 273.15 $$ We compare the result with the given options to find the correct answer. Thus, by solving the problem step-by-step using the ideal gas law, we can find the final temperature of the gas in the rigid tank after half of its mass has escaped.

Key concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In essence, it studies how thermal energy is converted to and from other forms of energy and how it affects matter. The fundamental concepts of thermodynamics are crucial for understanding various physical processes, including the behavior of gases within a closed system, like a rigid tank.

When dissecting a thermodynamic problem, such as the one presented in the exercise, we observe the principles of energy conversion. The filling and evacuation of gas from the tank are processes that can be analyzed using the laws of thermodynamics. Specifically, the first law, which is the principle of conservation of energy, must be satisfied; energy cannot be created or destroyed in an isolated system which influences the final state of the gas after the release from the tank.

Pressure Volume Temperature Relationship

The pressure volume temperature relationship, often encapsulated by the ideal gas law equation, expresses how these three state variables of a gas are interconnected. The ideal gas law is given by the equation PV = nRT. In this equation, P represents pressure, V is the volume, T is the temperature in Kelvin, n is the number of moles of the gas, and R is the universal gas constant.

The law is based on the assumption that the gas being analyzed behaves 'ideally,' meaning the molecules of the gas do not attract or repel each other and occupy no volume. Although no gas is perfectly ideal, many common gases exhibit behavior close enough to ideal conditions under a wide range of temperatures and pressures that the ideal gas law can be used with great accuracy.

Differentiating between absolute pressure (typically in atmospheres or kilopascals) and gauge pressure (which does not account for atmospheric pressure) is critical here. Similarly, temperature must always be in Kelvin for the formula to be correct, as Celsius does not accurately reflect the proportional relationships between volume and pressure.

Rigid Tank Gas Analysis

In the context of a rigid tank gas analysis, we apply the ideal gas law to a fixed-volume scenario. A rigid tank means that the volume, V, remains constant no matter what happens to the gas inside. Such containers are common in many practical thermodynamic applications like gas storage tanks, where the ability to withstand high pressures without deforming is critical.

In a rigid tank, when the amount of gas changes - through introduction or release of gas - the pressure and temperature will adjust, maintaining a relationship according to the ideal gas law. In the given exercise, when half of the gas mass is allowed to escape from the tank, we're essentially altering the number of moles (n) while keeping volume constant. By recalculating the moles based on the new mass and employing the ideal gas law, we can determine the new condition of the remaining gas, specifically the final temperature, as the problem requires.

In such problems, it's important to remember that we are dealing with an idealized situation. Real gases would require corrections for non-ideal behavior, particularly at high pressures or low temperatures, but in educational exercises the ideal gas law often suffices to illustrate the fundamental concepts of gas behavior in enclosed spaces.