Question

The gas inside a cylinder expands against a constant external pressure of 0.906 atm from a volume of 3.61 L to a volume of 14.50 L. In doing so, it turns a paddle immersed in 0.926 L of liquid octane (C8H18). Calculate the temperature rise of the liquid, assuming no loss of heat to the surroundings or frictional losses in the mechanism. Take the density of liquid C8H18 to be 0.703 g cm-3 and its specific heat to be 2.22 J °C-1 g-1.

Hints: The expansion work done by the gas is given by: w = -PexternalV and 1 L atm = 101.325 J.

(detla)T =
°C

Answer 1

expansion work done by the gas = - Pext X AV = -0.906 atm (14.50-3.61)L = -9.86634 L-atm = -9.86634 x 10)-325 J w = -999.707 J. (.- sign indicates work done by the gas). mass of liquid octane = 0.926 L x 1000 m² X 0.703 g - IL m = 650.978 9. 1 cm3

specific heat, 5= 2.22 g AT = change in Temp. heat absorbed by liquid = mxs x AT - work done by gas. 650.978 g x 2.22 J x ATC = 999.707 999.707 to AT = 650.976 fpx 2.23 7.8 AT = 0.63°c Answer: Temperature of liquid rises by 0.63°C