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part a
what is the pressure in a 19.0 l cylinder filled with 0.690 mol of nitrogen gas at 327 k?
pressure =
atm
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Step1: Recall Ideal Gas Law
The Ideal Gas Law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant (\( 0.0821 \, \text{L·atm/(mol·K)} \)), and \( T \) is temperature. We need to solve for \( P \), so rearrange the formula: \( P=\frac{nRT}{V} \).
Step2: Substitute Values
Given \( n = 0.690 \, \text{mol} \), \( R = 0.0821 \, \text{L·atm/(mol·K)} \), \( T = 327 \, \text{K} \), \( V = 19.0 \, \text{L} \). Plug into the formula:
\( P=\frac{0.690 \, \text{mol} \times 0.0821 \, \text{L·atm/(mol·K)} \times 327 \, \text{K}}{19.0 \, \text{L}} \).
First, calculate numerator: \( 0.690 \times 0.0821 \times 327 \approx 0.690 \times 26.8467 \approx 18.5242 \).
Then divide by volume: \( \frac{18.5242}{19.0} \approx 0.975 \).
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\( 0.975 \)