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.)


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

\[
PV = 8.31T
\]

Where:
- \( P \) is pressure in kilopascals (kPa),
- \( V \) is volume in liters (L),
- \( T \) 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, \( dP \), 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 \( PV = 8.31T \) with respect to time, we apply the product rule on the left-hand side:

\[
d(PV) = d(8.31T)
\]

This expands to:

\[
P \, dV + V \, dP = 8.31 \, dT
\]

We are solving for \( dP \), so rearrange the equation:

\[
V \, dP = 8.31 \, dT - P \, dV
\]

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

### Step 2: Determine Values for Substitution
Given:
- Initial volume: \( V = 14 \) L
- Final volume: \( V = 14.6 \) L
- \( dV = 14.6 - 14 = 0.6 \) L
- Initial temperature: \( T = 345 \) K
- Final temperature: \( T = 335 \) K
- \( dT = 335 - 345 = -10 \) K

### Step 3: Use the Ideal Gas Law to Find Initial Pressure
From the equation \( PV = 8.31T \), we can find the initial pressure, \( P \), when \( V = 14 \) L and \( T = 345 \) K:

\[
P \times 14 = 8.31 \times 345
\]

\[
P = \frac{8.31 \times 345}{14}
\]

Let's calculate this value.The initial pressure \( P \) is approximately \( 204.78 \, \text{kPa} \).

### Step 4: Substitute Values into the Differential Equation
Now that we have \( P = 204.78 \, \text{kPa} \), \( dV = 0.6 \, \text{L} \), and \( dT = -10 \, \text{K} \), we can substitute these into the equation:

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

Let's compute \( dP \).The approximate change in pressure, \( dP \), is \( -14.71 \, \text{kPa} \) (rounded to two decimal places). This indicates that the pressure decreases by about \( 14.71 \, \text{kPa} \) 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 \( 8.31 \) 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.

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.)

Find the approximate change in pressure when the volume of a gas increases from 14 L to 14.6 L and the temperature decreases from 345 K to 335 K using differentials. This problem involves the ideal gas law and uses concepts from differential calculus, providing a step-by-step solution for students in Grades 11-12.

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