What pressure (in kPa) does 2 mol of an ideal gas exert in a 20 L container at 250 K? (R = 8.314 J/(mol K))
Answer and explanation
Correct answer: C. 207.8 kPa
- Step 1: Ideal gas law: p = n R T / V with V in cubic meters.
- Step 2: p = 2 x 8.314 x 250 / 0.02 = 207850 Pa = 207.8 kPa.
Answer
207.8 kPa.
Why not the others: The volume must be converted from liters to cubic meters; the other values are off by a power of ten.
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