A 40.7mL trapped sample of Ar is at 0.9997 atm and 24.81°C. You increase the temperature to 42.73°C. What is?
A 40.7mL trapped sample of Ar is at 0.9997 atm and 24.81°C. You increase the temperature to 42.73°C. What is the final volume, in L, of your argon sample? What is the answer, right Sig Figs, and Units!
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use V1/T1 = V2/T2: with T = temperature in °K:
V2 = 40.7mL*(315.88/297.96) = 43.15 mL
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