Question

# The gas inside a cylinder expands against a constant external pressure of 0.943 atm from a...

The gas inside a cylinder expands against a constant external pressure of 0.943 atm from a volume of 3.35 L to a volume of 14.40 L. In doing so, it turns a paddle immersed in 0.951 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.

delta T = ________°C

Pext = 0.943 atm

Change in Volume = V2 - V1 = 14.40 - 3.35 = 11.05 L

thus, W = - pext * change in Volume = -0.943 atm * 11.05 L

= -10.42 L.atm * 101.325 J/L.atm = -1055.8J

Since no heat is lost

Thus, w = -q = -1055.8J

q = 1055.8J

Volume of octane = 0.951 L = 951 cm3

density of octane = 0.703 g/cm3

Mass of octane = Volume *density =  951 cm3​ * 0.703 g/cm3 = 668.553 g

specific heat = 2.22 J °C-1 g-1.

Now, q = m*specific heat * delta T

1055.8J = 668.553 g* 2.22 J °C-1 g-1.* delta T

delta T = 0.711 oC

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