Practice question
Question
A gas at 1.8atm and 27∘C occupies 3.5L. If the pressure is increased to 3atm and temperature raised to 127∘C, what is the percentage change in volume?
Explanation
Given: P1 = 1.8atm, T1 = 27∘C = 300K, V1 = 3.5L, P2 = 3atm, T2 = 127∘C = 400K. P1V1T1 = P2V2T2. V2 = V1×P1P2×T2T1 = 3.5×1.83×400300 = 3.5×0.6×1.3333≈2.8L. Percentage change: V1−V2V1×100 = 3.5−2.83.5×100 = 0.73.5×100 = 20% (decrease).