Answer to Question #86919 in General Chemistry for jason

Question

The gas inside a cylinder expands against a constant external pressure of 0.977 atm from a volume of 3.85 L to a volume of 14.60 L. In doing so, it turns a paddle immersed in 0.842 L of liquid benzyl alcohol (C7H7OH). 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 C7H7OH to be 1.04 g cm-3 and its specific heat to be 2.02 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.

T = _______ °C

Answer 1

Solution.

Q = delta U + A

Q(C7H7OH) = cm(delta T)

(delta T) = Q(C7H7OH) / cm

m = p*V

m = 1.04 g/cm3 *842 cm3 = 875.68 g

A = -Pexternal*(V2-V1) = -98944.525 Pa* (14.60 liters -3.85 liters ) = -1064191.144 J

delta U = 3/2 * p(V2-V1) = 3/2* 98944.525*(14.60-3.85) = 1596286.716 J

Q(C7H7OH) = 1596286.716 J - 1064191.144 J = 532095.572 J

(delta T) = Q(C7H7OH) / cm = 532095.572 / 2.02*875.68 = 300.81 oC

Answer:

(delta T) =  300.81 oC