Question

A \(3-m^{3}\) rigid tank contains hydrogen at 250 kPa and \(550 \mathrm{K} .\) The gas is now cooled until its temperature drops to \(350 \mathrm{K}\). Determine \((a)\) the final pressure in the tank and (b) the amount of heat transfer.

Short answer

Final Pressure: 158.96 kPa Amount of Heat Transfer: -232000 J (Heat is transferred out of the system)

Step-by-step solution

Write the given information and ideal gas law

The given information is: Initial temperature \(T_1=550K\), Final temperature \(T_2=350K\), Initial pressure \(P_1=250 kPa\) and volume \(V=3m^3\). We will use the ideal gas law: \(PV = nRT\) Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin.

Find the number of moles (n) of gas in the tank

Using the initial condition, we can find the number of moles (n) of gas using the ideal gas law as follows: \(250000 \times 3 = n \times 8.314 \times 550\) \(n = \frac{250000 \times 3}{8.314 \times 550}\) \(n = \approx 51.45 \thinspace moles\)

Calculate the final pressure (P2)

We will now use the ideal gas law again to determine the final pressure: \(P_2V = nRT_2\) \(P_2 = \frac{nRT_2}{V}\) \(P_2 = \frac{51.45 \times 8.314 \times 350}{3}\) \(P_2 = \approx 158960 \thinspace Pa\) \(P_2 = 158.96 \thinspace kPa\) (Final pressure)

Determine the initial and final internal energies

Since this is an ideal gas, we can use the following formula to determine the initial and final internal energies: \(U = \frac{3}{2}nRT\) \(U_1 = \frac{3}{2}(51.45)(8.314)(550) \quad => \quad U_1 \approx 625546 \thinspace J\) (Initial internal energy) and \(U_2 = \frac{3}{2}(51.45)(8.314)(350) \quad => \quad U_2 \approx 393546 \thinspace J\) (Final internal energy)

Calculate the amount of heat transfer (Q)

Finally, we can use the first law of thermodynamics to determine the amount of heat transfer: \(Q = \Delta U + W\) Since the process occurs inside a rigid tank, the work done (W) is zero. Thus, \(Q = U_2 - U_1\) \(Q = 393546 - 625546\) \(Q = -232000 \thinspace J\) (Amount of heat transfer) The negative sign indicates that the heat is transferred out of the system. To summarize, the final pressure in the tank is 158.96 kPa, and the amount of heat transfer is -232000 J.

Key concepts

Internal Energy

Internal energy is the sum of all the energy stored within a system. For ideal gases, this typically includes the kinetic and potential energies of the molecules. However, since potential energy due to molecular interactions is negligible for ideal gases, internal energy largely depends on kinetic energy and, consequently, on the temperature of the gas. An important thing to remember is that internal energy is a state function, meaning it only depends on the current state of the system and not on how it got there.

In the solved exercise, the internal energy changes when the gas cools down. The formula \(U = \frac{3}{2}nRT\) beautifully illustrates that as the temperature (T) decreases, so does the internal energy (U), because the random movement of particles slows down.

First Law of Thermodynamics

The first law of thermodynamics is a version of the law of conservation of energy and states that the energy of an isolated system is constant. Energy within a system can change forms, such as from kinetic to potential, but the total amount remains constant. The formula for the first law is \(Q = \Delta U + W\), where Q is the heat exchanged, \(\Delta U\) is the change in internal energy, and W is the work done by or on the system.

In our exercise, the tank is rigid, meaning it does no work as its volume does not change (W = 0). Therefore, the heat transferred (\(Q\)) is equal to the change in internal energy (\(\Delta U\)), which is aligned with the mentioned principle.

Heat Transfer

Heat transfer is the process of energy moving from one body or system to another due to a temperature difference. There are three main modes of heat transfer: conduction, convection, and radiation. In thermodynamics, heat is often symbolized by Q and is measured in Joules.

The step by step solution shows heat is lost from the system (\(Q < 0\)), because the internal energy decreases. This demonstrates the natural tendency for heat to flow from a hotter to a cooler place, in this case from the gas to its surroundings affecting the internal energy directly.

Rigid Tank Thermodynamics

In thermodynamics, a rigid tank is a system with a fixed volume. No work is done when the contents of a rigid tank undergo a process since work in thermodynamics is often a result of volume change (\(W=P\Delta V\)), and in a rigid tank, \(\Delta V=0\). This simplifies calculations as the work term in the first law of thermodynamics can be disregarded.

The problem presented involves a rigid tank where hydrogen gas is cooled. As per the results of the step by step solution, the final pressure of the gas within can be calculated using the ideal gas law, and the heat transfer can be found by analyzing the change in internal energy without accounting for work.