Question

An ideal gas (X) present in a vessel of volume \( V \) exerted a pressure of 16.4 atm at 200 K. What is its concentration in mol \( L^{-1} \)? % Given Data Given \( R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1} \)

• A. \( 0.50 \, \text{mol L}^{-1} \)
• B. \( 0.25 \, \text{mol L}^{-1} \)
• C. \( 1.00 \, \text{mol L}^{-1} \)
• D. \( 1.50 \, \text{mol L}^{-1} \)

Hint:

Always remember that concentration is simply \( \frac{n}{V} \), which is moles per unit volume. In problems involving ideal gases, use the ideal gas law to find the concentration.

Answer 1

The Correct Option is C

We can use the ideal gas law, which is given by: \[ PV = nRT \] Where: - \( P \) is the pressure (16.4 atm) - \( V \) is the volume (given as \( V \) L) - \( n \) is the number of moles - \( R \) is the ideal gas constant (0.082 L atm mol\(^{-1}\) K\(^{-1}\)) - \( T \) is the temperature (200 K) We can rewrite the equation to find the concentration (\( n/V \)): \[ P = \frac{nRT}{V} \] \[ \frac{n}{V} = \frac{P}{RT} \] Substitute the known values: \[ \frac{n}{V} = \frac{16.4 \, \text{atm}}{(0.082 \, \text{L atm mol}^{-1} \text{K}^{-1}) \times (200 \, \text{K})} \] \[ \frac{n}{V} = \frac{16.4}{16.4} = 1.00 \, \text{mol L}^{-1} \] Thus, the concentration of the gas is \( 1.00 \, \text{mol L}^{-1} \).

Answer 2

Step 1: Recall the ideal gas equation.
\[ PV = nRT, \] where
\(P\) = pressure (atm),
\(V\) = volume (L),
\(n\) = number of moles,
\(R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1}\) (ideal gas constant),
\(T\) = temperature (K).

Step 2: Express concentration.
Concentration \(C = \frac{n}{V}\) (moles per liter).
Rearranging the ideal gas equation,
\[ n = \frac{PV}{RT} \implies C = \frac{n}{V} = \frac{P}{RT}. \]

Step 3: Substitute given values.
\[ C = \frac{16.4 \, \text{atm}}{0.082 \times 200} = \frac{16.4}{16.4} = 1.00 \, \text{mol L}^{-1}. \]

Step 4: Conclusion.
The concentration of the ideal gas \(X\) in the vessel is \(\boxed{1.00 \, \text{mol L}^{-1}}\).