Question 9 of 22 What pressure is exerted by 0.660 mol N, in a 8.84 L steel container at 48.7 °C? P = atm

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Chapter1: Chemical Foundations
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**Question:**

What pressure is exerted by 0.660 mol of \( N_2 \) in an 8.84 L steel container at 48.7 °C?

**Answer:**

\( P = \) [Pressure in atm]

**Explanation:**

To find the pressure exerted by the nitrogen gas (\( N_2 \)), you can use the ideal gas law equation:

\[ 
PV = nRT 
\]

where:
- \( P \) is the pressure in atmospheres (atm).
- \( V \) is the volume in liters (L), which is 8.84 L.
- \( n \) is the number of moles, which is 0.660 mol.
- \( R \) is the ideal gas constant, 0.0821 L·atm/mol·K.
- \( T \) is the temperature in Kelvin (K).

First, convert the temperature from Celsius to Kelvin:

\[
T(K) = 48.7 + 273.15 = 321.85 \, K
\]

Using the ideal gas law, rearrange for pressure \( P \):

\[
P = \frac{nRT}{V} = \frac{(0.660 \, \text{mol})(0.0821 \, \text{L·atm/mol·K})(321.85 \, \text{K})}{8.84 \, \text{L}}
\]

Calculate the result to find the pressure in atm.
Transcribed Image Text:**Question:** What pressure is exerted by 0.660 mol of \( N_2 \) in an 8.84 L steel container at 48.7 °C? **Answer:** \( P = \) [Pressure in atm] **Explanation:** To find the pressure exerted by the nitrogen gas (\( N_2 \)), you can use the ideal gas law equation: \[ PV = nRT \] where: - \( P \) is the pressure in atmospheres (atm). - \( V \) is the volume in liters (L), which is 8.84 L. - \( n \) is the number of moles, which is 0.660 mol. - \( R \) is the ideal gas constant, 0.0821 L·atm/mol·K. - \( T \) is the temperature in Kelvin (K). First, convert the temperature from Celsius to Kelvin: \[ T(K) = 48.7 + 273.15 = 321.85 \, K \] Using the ideal gas law, rearrange for pressure \( P \): \[ P = \frac{nRT}{V} = \frac{(0.660 \, \text{mol})(0.0821 \, \text{L·atm/mol·K})(321.85 \, \text{K})}{8.84 \, \text{L}} \] Calculate the result to find the pressure in atm.
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