A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at 320 K, the volume of the upper

Question

A vertical cylinder closed from both ends is equipped with an easily moving piston dividing the volume into two parts, each containing one mole of air. In equilibrium at 320 K, the volume of the upper part is 4.0 times greater than that of the lower part. At what temperature (in kelvin) will the ratio of these volumes be equal to 3.0?

TestHub · Accepted Answer

Equal moles and temperature on both sides, so V_u/V_l = P_l/P_u. Piston weight gives P_l = P_u + mg/S (constant difference).

At 320 K: P_l/P_u = 4 ⇒ mg/S = 3P_u. Total volume C: V_u = 4C/5, V_l = C/5, and P_u = R × 320/(4C/5) = 400R/C.

At T: ratio 3 ⇒ P_l' = 3P_u', and P_l' - P_u' = 2P_u' = mg/S = 3P_u ⇒ P_u' = 1.5P_u.

Also P_u' = RT'/(3C/4) = 4RT'/3C. So 4T'/(3C) = 1.5 × 400/C ⇒ 4T'/3 = 600 ⇒ T' = 450 K ⇒ 0450.

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