Question

At \( T(K) \), three moles of an ideal gas is present in a 10 L vessel. If the kinetic energy of an ideal gas is \( 3000 \) J mol\(^{-1}\), the approximate pressure of the gas (in atm) is:

• A. \( 59.2 \)
• B. \( 5.92 \)
• C. \( 0.592 \)
• D. \( 11.84 \)

Hint:

- The kinetic energy formula is directly linked to temperature via \( KE = \frac{3}{2} RT \). - Always use the appropriate value of \( R \) based on the required units (J/mol-K for energy, atm L/mol-K for gas equations).

Answer 1

The Correct Option is B

Step 1: Using the relation for kinetic energy The kinetic energy per mole of an ideal gas is given by: \[ KE = \frac{3}{2} RT \] Given, \( KE = 3000 \) J/mol, we solve for \( T \): \[ T = \frac{2 \times 3000}{3R} \] Using \( R = 8.314 \) J/mol-K: \[ T = \frac{6000}{3 \times 8.314} = \frac{6000}{24.942} \approx 240.6 K \] Step 2: Applying the Ideal Gas Law The ideal gas equation is: \[ PV = nRT \] Given: \[ n = 3 \text{ moles}, \quad V = 10 \text{ L}, \quad R = 0.0821 \text{ atm L/mol K}, \quad T = 240.6 \text{ K} \] \[ P \times 10 = 3 \times 0.0821 \times 240.6 \] \[ P = \frac{3 \times 0.0821 \times 240.6}{10} \] \[ P = \frac{59.2}{10} = 5.92 \text{ atm} \]