Problem 38

Question

The pressure, volume, and temperature of a mole of an ideal gas are related by the equation $P V=8.31 T,$ where $P$ is measured in kilo pascals, $V$ in liters, and $T$ in kelvins. Use differentials to find the approximate change in the pressure if the volume increases from 12 L to 12.3 $\mathrm{L}$ and the temperature decreases from 310 $\mathrm{K}$ to 305 $\mathrm{K}$ .

Step-by-Step Solution

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Answer

The pressure decreases by approximately 8.82 kPa.

1Step 1: Understanding the equation

The given equation is the ideal gas law for one mole, expressed as $ PV = 8.31 T $. This signifies the relationship between pressure $ P $, volume $ V $, and temperature $ T $. We want to find the change in pressure $ dP $ using differentials.

2Step 2: Differentiating the equation

Taking the differential of both sides of the equation $ PV = 8.31 T $, we apply the product rule for the left-hand side: $ PdV + VdP = 8.31 dT $. This equation will help us calculate $ dP $.

3Step 3: Substituting the initial values

Initially, $ V_0 = 12 $ L and $ T_0 = 310 $ K. We calculate the initial pressure $ P_0 $ using $ P_0 V_0 = 8.31 T_0 $. Substituting the values, $ P_0 \times 12 = 8.31 \times 310 $, gives us $ P_0 = \frac{8.31 \times 310}{12} \approx 214.38 $ kPa.

4Step 4: Calculating changes in variables

Given $ dV = 12.3 - 12 = 0.3 $ L and $ dT = 305 - 310 = -5 $ K, use these in the differential equation $ PdV + VdP = 8.31 dT $. With $ P_0 = 214.38 $ and $ V_0 = 12 $, we substitute these values into the differential equation.

5Step 5: Solving for the change in pressure

Substitute the known values: $ 214.38 \times 0.3 + 12 \times dP = 8.31 \times (-5) $. Simplifying, we get $ 64.314 + 12dP = -41.55 $. Solving for $ dP $, we find $ 12dP = -41.55 - 64.314 = -105.864 $, hence $ dP = \frac{-105.864}{12} \approx -8.822 $ kPa.

Key Concepts

Ideal Gas LawPressure-Volume-Temperature RelationshipProduct Rule in Differentiation

Ideal Gas Law

The Ideal Gas Law is a fundamental concept in chemistry and physics. It describes the behavior of gases under different conditions. The law is usually stated as $ PV = nRT $. However, in our problem, since we're dealing with one mole of gas, it simplifies to $ PV = 8.31T $. Here, $ P $ represents pressure, $ V $ is volume, and $ T $ is temperature, with $ R $ being the gas constant given as 8.31 when dealing with specific units.
The Ideal Gas Law helps us understand how gases respond to changes in pressure, volume, and temperature. It's an approximation that assumes no interactions between gas molecules and that the volume of the molecules themselves is negligible. This is true under normal conditions but becomes less accurate at high pressures or low temperatures.

Pressure (P): Measured in kilopascals (kPa).
Volume (V): Measured in liters (L).
Temperature (T): Measured in kelvins (K).

Understanding this equation is crucial for solving problems related to gas behaviors, such as finding changes in pressure, volume, or temperature.

Pressure-Volume-Temperature Relationship

The pressure-volume-temperature relationship is how pressure, volume, and temperature influence one another in a gas. In our problem, we investigate how changes in volume and temperature affect pressure. This is crucial for various scientific and engineering applications.
When temperature changes, it affects how fast the molecules of a gas move. If temperature decreases (cooling), the energy of the molecules drops, causing them to collide with the walls of the container less frequently, thus reducing pressure if volume is constant.
On the other hand, when volume increases, there's more space for the gas particles to move around. This tends to decrease the pressure as well, assuming temperature remains constant. The gas law equation ($ PV = 8.31 T $) ties these concepts together, showing their interdependence.

Increasing volume leads to decreasing pressure if temperature doesn’t change.
Decreasing temperature lowers pressure if volume is held constant.

In our example, we used the differential form of this equation, allowing us to approximate small changes in pressure given changes in volume ($ dV $) and temperature ($ dT $). This approximation helps in practical scenarios where exact changes are complex to compute.

Product Rule in Differentiation

The product rule is a vital tool in calculus, especially when dealing with more than one variable. It was necessary to apply this rule to find the differential form of the ideal gas law. Here's a simple explanation of how the product rule works: if you have a product of two functions, say $ u(x) $ and $ v(x) $, then the derivative of their product is given by the formula $ (uv)' = u'v + uv' $.
In the context of our exercise, the left side of the equation, $ PV $, is a product of two variables, pressure $ P $ and volume $ V $. To differentiate $ PV $ with respect to time, we apply the product rule: $ PdV + VdP = 8.31dT $. Here is how each term relates:

$ PdV $ represents the change in volume while keeping pressure constant.
$ VdP $ represents the change in pressure while keeping volume constant.

This rule enables us to isolate $ dP $, the change in pressure, given changes in the other parameters. Applying the product rule is fundamental to manipulating the differential form of physical laws, making it a powerful technique in scientific calculations.

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