A 1.00 mol sample of an ideal gas occupies 2.00 L at 298 K. What is the pressure? (R = 0.0821 L atm mol^-1 K^-1)

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Multiple Choice

A 1.00 mol sample of an ideal gas occupies 2.00 L at 298 K. What is the pressure? (R = 0.0821 L atm mol^-1 K^-1)

Explanation:
The pressure is found using PV = nRT, solving for P: P = nRT / V. With n = 1.00 mol, R = 0.0821 L atm mol^-1 K^-1, T = 298 K, and V = 2.00 L, compute nRT = 1.00 × 0.0821 × 298 ≈ 24.50 L·atm. Divide by V: 24.50 / 2.00 ≈ 12.25 atm. So the pressure is about 12.2 atm. The other values arise from arithmetic mistakes, such as not dividing by the correct volume or misplacing the decimal.

The pressure is found using PV = nRT, solving for P: P = nRT / V. With n = 1.00 mol, R = 0.0821 L atm mol^-1 K^-1, T = 298 K, and V = 2.00 L, compute nRT = 1.00 × 0.0821 × 298 ≈ 24.50 L·atm. Divide by V: 24.50 / 2.00 ≈ 12.25 atm. So the pressure is about 12.2 atm. The other values arise from arithmetic mistakes, such as not dividing by the correct volume or misplacing the decimal.

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