Answer to Question #235104 in Molecular Physics | Thermodynamics for Unknown346307

Question

. A 15 cm diameter vertical cylinder, closed by a piston contains a combus￾tible mixture at a temperature of 30°C. The piston is free to move and its weight is such that the

mixture pressure is 3 bar. Upper surface of the piston is exposed to the atmosphere. The mixture

is ignited. As the reaction proceeds, the piston moves slowly upwards and heat transfer to the

surroundings takes place. When the reaction is complete and the contents have been reduced to

the initial temperature of 30°C, it is found that the piston has moved upwards a distance of

8.5 cm and the magnitude of heat transfer is 4 kJ. Evaluate :

(i) The work ;

(ii) Decrease in internal energy of the system

Answer 1

Diameter of vertical cylinder, d = 15 cm (or 0.15 m)

Temperature of combustible mixture = 30°C (or 303 K)

Pressure of the mixture = 3 bar $= 3 \times 10^5 \; N/m^2$

Upward displacement of the system = 8.5 cm (or 0.085 m)

Magnitude of heat transfer, Q = – 4 kJ

(i) Work done by the system, $W = \int pdv$

$p = 3 \times 10^5 \;N/m^2 \\ W= 3 \times 10^5 \int dv \\ = 3 \times 10^5 [\frac{\pi}{4} \times (0.15)^2 \times 0.085] \;N-m \\ = 450.62 \;N-m \\ = 0.4506 \;kJ \\ W= 0.4506 \;kJ$

(ii) By first law of thermodynamics,

$Q=ΔU+W \\ -4 = ΔU + 0.4506 \\ ΔU = -4.4506 \;kJ$

Decrease in internal energy = 4.4506 kJ.