State, in words, the relation between the increase in the internal energy of a gas, the work done on the gas, and the heat supplied to the gas.
(a) A quantity of 0.200 mol of air enters a diesel engine at a pressure of 1.04 * 105 Pa and at a temperature of 297 K. Assuming that air behaves as an ideal gas, find the volume of this quantity of air.
(b) The air is then compressed to one twentieth of this volume, the pressure having risen to 6.89 * 105 Pa. Find the new temperature.
(c) Heating of the air then takes place by burning a small quantity of fuel in it to supply 6150 J. This is done at a constant pressure of 6.89 * 105 Pa as the volume of air increases and the temperature rises to 2040 K. Find
(i) the volume of the gas after burning the fuel,
(ii) the work done by the air during this expansion,
(iii) the change in the internal energy of the air during this expansion.
Answer:
The increase in internal energy of a gas is equal to the sum of the work done on the gas and the heat supplied to the gas.
(a) PV = nRT
V = nRT/P
= (0.200 * 8.31 * 297) / (1.04 * 105)
= 4.75 * 10-3 m3
(b) Given V' = V/20, and P' = 6.89 * 105 Pa,
= 98.4 K
(c)
(i) T1/T2 = V1 / V2 (when pressure is constant)
V2 = 2040 / 98.4 * V'
V2 = 2040 / 98.4 * 2.37 * 10-4
= 4.92 * 10-3 m3
(ii) Work done by air
= p ΔV
= 6.89 * 105 * (4.92 - 0.237) * 10-3
= 3230 J
(iii) ΔU = Q supplied + WD on air
= Q supplied - WD by air
= 6150 - 3230
= 2920 J
