A cylinder with a movable piston contains 2.00g of helium, , at room temperature

Question

A cylinder with a movable piston contains 2.00g of helium, , at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70L ? (The temperature was held constant.)

Explanation / Answer

P x V = n x R x T n = m/MW ==> P x V = (m/MW) x R x T Solving for (R / MW) ==> R / MW = (P x V) / (m x T) R / MW = constant for all states So, at states 1 and 2, (P1 x V1) / (m1 x T1) = (P2 x V2) / (m2 x T2) Given: P1 = P2 and T1 = T2 (both constant) ==> V1 / m1 = V2 / m2 Solving for m2, m2 = (V2 x m1) / V1 Given: V1 = 2.00L m1 = 2.00 g V2 = 2.7L Substituting values, m2 = (2.7L x 2.00 g) / 2.00L = 2.5 g The mass added = m2 - m1 = 2.5 g - 2.00 g = 0.5 g of He

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