Problem 8 - Active Calculus Multivariable

Question

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy \(U\) (in Joules) of the ammonia is a function of the volume \(V\) (in cubic meters) of the container, and the temperature \(T\) (in degrees Kelvin) of the gas. The differential \(d U\) is given by \(d U=840 d V+27.32 d T\). (a) How does the energy change if the volume is held constant and the temperature is decreased slightly? \(\odot\) it increases slightly \(\odot\) it does not change \(\odot\) it decreases slightly (b) How does the energy change if the temperature is held constant and the volume is increased slightly? \(\odot\) it does not change \(\odot\) it increases slightly \(\odot\) it decreases slightly (c) Find the approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin. Change in energy = ___________. Please include units in your answer.

Step-by-Step Solution

Answer

(a) The energy decreases slightly. (b) The energy increases slightly. (c) The approximate change in energy is 81.834 Joules.

Step 1: Analyzing the given formula

The given formula in the exercise is \(d U = 840 dV + 27.32 dT\). (a) If the volume is held constant and the temperature is decreased slightly:

Step 2: Calculating change in energy while holding volume constant

When the volume is held constant, \(dV = 0\). Now, let's plug this into the given formula: \(dU = 840(0) + 27.32 dT\). If the temperature is decreased slightly, \(dT\) will be negative. So, the change in energy \(dU\) will also be negative. The correct answer is: \(\odot\) it decreases slightly. (b) If the temperature is held constant and the volume is increased slightly:

Step 3: Calculating change in energy while holding temperature constant

When the temperature is held constant, \(dT = 0\). Now, let's plug this into the given formula: \(dU = 840 dV + 27.32(0)\). If the volume is increased slightly, \(dV\) will be positive. So, the change in energy \(dU\) will also be positive. The correct answer is: \(\odot\) it increases slightly. (c) If the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin:

Step 4: Converting cubic centimeters to cubic meters

We need to convert 150 cubic centimeters to cubic meters. 1 cubic meter = 1000000 cubic centimeters. So, \(150 \text{ cm}^3 \times \frac{1 \text{ m}^3}{1000000 \text{ cm}^3} = 0.00015 \text{ m}^3\).

Step 5: Calculating change in energy

Now that we have the changes in volume (\(-0.00015\) m³) and temperature (+3 K), we can plug these values into the given formula: \(\Delta U = 840(-0.00015) + 27.32(3)\) \(\Delta U = -0.126 + 81.96\) \(\Delta U \approx 81.834 \text{ Joules}\). The approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin is 81.834 Joules.

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that deals with the calculation of derivatives. In the context of the ammonia gas problem, differential calculus allows us to understand how small changes in volume (\(dV\)) and temperature (\(dT\)) affect the total energy (\(U\)) of the system. The differential form \(dU = 840 dV + 27.32 dT\) represents the rate at which the energy changes with respect to small changes in volume and temperature, known as the first-order derivatives.

These derivatives indicate the sensitivity of the energy to the respective variables. For example, the coefficient 840 in front of \(dV\) shows the amount by which energy increases with each cubic meter increase in volume, assuming temperature is constant. Similarly, 27.32 represents the energy change per degree Kelvin increase in temperature, assuming constant volume. By using differentials, we can calculate the infinitesimal changes in energy for infinitely small changes in volume and temperature, which is a powerful tool in engineering and physics.

Energy Change in Physical Systems

Energy change in physical systems is often dependent on various factors, such as temperature and volume for a gas contained in a vessel. With thermodynamics, we quantify these changes to understand how physical systems behave under different conditions. In the given exercise, we observe how energy (\(U\)) changes with different scenarios.

In scenario (a), when volume is held constant and temperature decreases, we see that the energy decreases slightly. This is because energy is directly proportional to temperature in many physical systems. In contrast, in scenario (b), increasing volume with constant temperature results in an increase in energy. This is characteristic of gases, which typically require work to be done on them in order to increase volume, hence increasing energy. Furthermore, applying the given formula allows us to mathematically determine the exact change in energy, making these abstract concepts tangible in practicable terms.

Partial Derivatives

Partial derivatives play a pivotal role when dealing with functions that depend on more than one variable, like the energy function \(U(V, T)\) for ammonia gas. In our exercise, the differential \(dU\) demonstrates how the energy changes through partial derivatives with respect to volume and temperature. When we hold volume constant (\(dV=0\)), we are essentially taking the partial derivative of \(U\) with respect to temperature (\(T\)).

This shows us how the energy will change with a slight decrease in temperature, independent of the volume. Conversely, holding temperature constant (\(dT=0\)) and finding how \(U\) changes with volume is taking the partial derivative with respect to volume (\(V\)). It is through the consideration of partial derivatives that we are able to isolate and examine the impact of changing one variable while holding others constant, which is a common situation in physical and engineering problems.