Math Problem Statement

The pressure, volume, and temperature of a mole of an ideal gas are related by the equation PV = 8.31T, where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approximate change in the pressure (in kPa) if the volume increases from 14 L to 14.6 L and the temperature decreases from 345 K to 335 K. (Note whether the change is positive or negative in your answer. Round your answer to two decimal places.)

Solution

We are given the ideal gas law equation for a mole of an ideal gas:

PV=8.31TPV = 8.31T

Where:

  • PP is pressure in kilopascals (kPa),
  • VV is volume in liters (L),
  • TT is temperature in kelvins (K),
  • 8.31 is the gas constant (in kPa·L/mol·K).

We are tasked with finding the approximate change in pressure, dPdP, when the volume increases from 14 L to 14.6 L, and the temperature decreases from 345 K to 335 K, using differentials.

Step 1: Implicit Differentiation

Differentiating both sides of the equation PV=8.31TPV = 8.31T with respect to time, we apply the product rule on the left-hand side:

d(PV)=d(8.31T)d(PV) = d(8.31T)

This expands to:

PdV+VdP=8.31dTP \, dV + V \, dP = 8.31 \, dT

We are solving for dPdP, so rearrange the equation:

VdP=8.31dTPdVV \, dP = 8.31 \, dT - P \, dV

dP=8.31dTPdVVdP = \frac{8.31 \, dT - P \, dV}{V}

Step 2: Determine Values for Substitution

Given:

  • Initial volume: V=14V = 14 L
  • Final volume: V=14.6V = 14.6 L
  • dV=14.614=0.6dV = 14.6 - 14 = 0.6 L
  • Initial temperature: T=345T = 345 K
  • Final temperature: T=335T = 335 K
  • dT=335345=10dT = 335 - 345 = -10 K

Step 3: Use the Ideal Gas Law to Find Initial Pressure

From the equation PV=8.31TPV = 8.31T, we can find the initial pressure, PP, when V=14V = 14 L and T=345T = 345 K:

P×14=8.31×345P \times 14 = 8.31 \times 345

P=8.31×34514P = \frac{8.31 \times 345}{14}

Let's calculate this value.The initial pressure PP is approximately 204.78kPa204.78 \, \text{kPa}.

Step 4: Substitute Values into the Differential Equation

Now that we have P=204.78kPaP = 204.78 \, \text{kPa}, dV=0.6LdV = 0.6 \, \text{L}, and dT=10KdT = -10 \, \text{K}, we can substitute these into the equation:

dP=8.31(10)204.78×0.614dP = \frac{8.31 \, (-10) - 204.78 \times 0.6}{14}

Let's compute dPdP.The approximate change in pressure, dPdP, is 14.71kPa-14.71 \, \text{kPa} (rounded to two decimal places). This indicates that the pressure decreases by about 14.71kPa14.71 \, \text{kPa} due to the volume increase and temperature decrease.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

  1. How would the change in pressure differ if the temperature remained constant?
  2. What is the impact on pressure if the volume were to decrease instead of increase?
  3. How does the gas constant 8.318.31 change for gases other than ideal gases?
  4. Can you use differentials to estimate changes in other properties of gases, like internal energy?
  5. How does this result compare with an actual physical experiment involving real gases?

Tip: In thermodynamics, using differentials helps estimate small changes efficiently, but for larger changes, more precise methods may be needed.

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Math Problem Analysis

Mathematical Concepts

Differential Calculus
Physics
Thermodynamics

Formulas

Ideal Gas Law: PV = 8.31T
Differential form: V dP = 8.31 dT - P dV

Theorems

Product Rule in Differentiation

Suitable Grade Level

Grades 11-12 (Advanced High School / Early College)