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what is the volume, in l, of 6.54 moles of an ideal gas at 27.9°c and 2…

Question

what is the volume, in l, of 6.54 moles of an ideal gas at 27.9°c and 2.06 atm? report your answer to one place after the decimal point.
r = 0.0821 \frac{l\cdot atm}{mol\cdot k} or 8.314 \frac{kpa\cdot l}{mol\cdot k}

Explanation:

Step1: Convert temperature to Kelvin

To convert Celsius to Kelvin, we use the formula \( T(K) = T(^\circ C) + 273.15 \). Given \( T = 27.9^\circ C \), so \( T = 27.9 + 273.15 = 301.05 \, K \).

Step2: Use the ideal gas law formula

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant, and \( T \) is temperature. We need to solve for \( V \), so rearrange the formula to \( V=\frac{nRT}{P} \).

Given:

  • \( n = 6.54 \, mol \)
  • \( R = 0.0821 \frac{L\cdot atm}{mol\cdot K} \) (since pressure is in atm)
  • \( T = 301.05 \, K \)
  • \( P = 2.06 \, atm \)

Substitute the values into the formula:

\( V=\frac{6.54 \, mol\times0.0821 \frac{L\cdot atm}{mol\cdot K}\times301.05 \, K}{2.06 \, atm} \)

Step3: Calculate the numerator first

Calculate \( nRT \):

\( 6.54\times0.0821\times301.05 \)

First, \( 6.54\times0.0821 \approx 0.537 \)

Then, \( 0.537\times301.05 \approx 161.67 \)

Step4: Divide by pressure

Now divide by \( P = 2.06 \, atm \):

\( V=\frac{161.67}{2.06} \approx 78.5 \, L \)

Answer:

78.5