Question

The gas inside a cylinder expands against a constant external pressure of 0.953 atm from a volume of 4.51 L to a volume of 14.50 L. In doing so, it turns a paddle immersed in 0.962 L of liquid toluene (C7H8). 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 C7H8 to be 0.867 g cm-3 and its specific heat to be 1.69 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.

Answer 1

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9 - Amou leat TW = - Pexternal AV I given IL. alin = 101.325T Pexternal = 0.953 atm. AV=v-v. = 14.50-wash -9.52 l. ahm = ? ? J = 9.994 -9.sak.ahm x 101.325] W =- (0.952 atm) (IOL) 1.k.al W = -9.52b.alm = -964.625 Here, work used to transfer to liquid as Now, use q = m. Co.AT heat q = ove (heat released) Then, heat absorbed in liquid so q=tve leased / absorbed by the substance = +965.625 m= mass of substance density = mass modxv. volume Cp = heat capacity of substance AT = change in temperature Cemicsaluce = 1.6938gy m= (0.8679.cm) 6-9621) - (0.867 g/cm3 (962 mL) = (0.8679/613 (962 Curs) +964.623 - 834.0549 (834.0543)(16936j) = +964 62 °C = +0. 68°c 1409.55 8. The temperature will raise by 0.68°C → AT = a.co