For an ideal gas, calculate the pressure when 0.50 mol gas occupies 12.0 L at 298 K.

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Multiple Choice

For an ideal gas, calculate the pressure when 0.50 mol gas occupies 12.0 L at 298 K.

Explanation:
This uses the ideal gas law, P = nRT / V. With n = 0.50 mol, T = 298 K, V = 12.0 L, and R = 0.082057 L·atm/(mol·K), the pressure is P = (0.50)(0.082057)(298) / 12.0 ≈ 1.02 atm. Think of it this way: at 298 K, 1 mole of an ideal gas would occupy about 24.45 L at 1 atm. Half a mole in 12 L gives the same pressure as about 1 atm. The other numbers would require either a much smaller volume (to reach higher pressures) or a larger volume (to drop pressure), or a different temperature.

This uses the ideal gas law, P = nRT / V. With n = 0.50 mol, T = 298 K, V = 12.0 L, and R = 0.082057 L·atm/(mol·K), the pressure is P = (0.50)(0.082057)(298) / 12.0 ≈ 1.02 atm.

Think of it this way: at 298 K, 1 mole of an ideal gas would occupy about 24.45 L at 1 atm. Half a mole in 12 L gives the same pressure as about 1 atm. The other numbers would require either a much smaller volume (to reach higher pressures) or a larger volume (to drop pressure), or a different temperature.

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