Question

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 = -P_externalV and 1 L atm = 101.325 J.

delta T = ________°C

Answer 1

P_ext = 0.943 atm

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

thus, W = - p_ext * 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