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solve this problem. Hint: You will need to use the ideal gas law to solve this problem.

To solve this problem, we can use the **Ideal Gas Law**:

\[
PV = nRT
\]

where:
- \( P \) is the pressure (in atm),
- \( V \) is the volume (in liters),
- \( n \) is the amount of substance (in moles),
- \( R \) is the gas constant (0.0821 atm·L/mol·K),
- \( T \) is the temperature (in Kelvin).

### Given values:
- \( P = 1.87 \, \text{atm} \)
- \( V = 45.9 \, \text{L} \)
- \( n = 8.15 \, \text{mol} \)
- \( R = 0.0821 \, \text{atm·L/mol·K} \)

We are asked to find the temperature \( T \).

### Rearranging the Ideal Gas Law
To find \( T \), rearrange the formula:

\[
T = \frac{PV}{nR}
\]

### Plugging in the values:
\[
T = \frac{(1.87 \, \text{atm}) \times (45.9 \, \text{L})}{(8.15 \, \text{mol}) \times (0.0821 \, \text{atm·L/mol·K})}
\]

Calculating this:

1. Calculate the numerator:
   \[
   1.87 \times 45.9 = 85.893
   \]

2. Calculate the denominator:
   \[
   8.15 \times 0.0821 = 0.669215
   \]

3. Now, divide the results:
   \[
   T = \frac{85.893}{0.669215} \approx 128.37 \, \text{K}
   \]

### Answer:
The temperature \( T \) is approximately **128.37 K**.

Would you like more details or have any questions?

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Here are five related questions for further exploration:

1. How would the temperature change if the volume of the gas were increased?
2. What would happen to the temperature if the pressure increased while keeping the volume constant?
3. How does the ideal gas law differ for real gases at high pressures?
4. What would be the temperature if \( R \) were given in different units, such as J/(mol·K)?
5. How can the ideal gas law be used to calculate molar mass?

**Tip:** Always ensure that all units are compatible when using the ideal gas law (atm, L, mol, and K).

If 8.15 mol of a gas has a volume of 45.9 L and a pressure of 1.87 atm, what is the temperature? Use R = 0.0821 atm·L/mol·K.

Learn how to calculate the temperature of a gas using the Ideal Gas Law, given 8.15 mol, 45.9 L volume, and 1.87 atm pressure. This detailed solution includes key steps and considerations for students in Grades 10-12.

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