Question

THE PRESSURE OF GAS CHANGES LINEARLY WITH VOLUME 10 Kpa , 200 cm³ to 50 Kpa, 50 cm³ .
a) calculate the work done by the gas.
b) if no heat is supplied or extracted from the gas, what is the change in the internal energy?

please answer fast. ​

Answer 1

Sol. P base 1 = 10 kpa = 10 × 10^3 pa. P base 2 = 50 × 10^3 pa. v base 1 = 200 cc. v base 2 = 50 cc (i) Work done on the gas = ½ (10+50) x 10^3 x (50-200) x 10 ^-6 = - 4.5 J (ii) dQ = 0 ⇒ 0 = du + dw ⇒ du = – dw = 4.5 J

Answer 2

Initial pressure of the system, P1 = 10 kPa = 10 × 103 Pa

Final pressure of the system, P2 = 50 kPa = 50 × 103 Pa

Initial volume of the system, V1 = 200 cc

Final volume of the system, V2 = 50 cc

(i) Work done on the gas = Pressure × Change in volume of the system

Since pressure is also changing, we take the average of the given two pressures.

Now,

P = 1210+50×103

  = 30×103 Pa

Work done by the system of gas can be given by

30×103×50-200×10-6=-4.5 J

(ii) Since no heat is supplied to the system, ∆Q = 0.

Using the first law of thermodynamics, we get

∆U = − ∆W = 4.5 J

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