Question

Three kilograms of argon (Ar) changes from an initial volume and a temperature of 298K to (a) four times the volume and a temperature of 298K and (b) one fourth the original volume under adiabatic conditions. In each case: Calculate the amount of work performed. Is the work done by the system or the surroundings? (pl/, for monatomic gases is 1.667)

Answer 1

Answer:

Explanation:

3 kg = 3 / 40 = .075 moles = n

a ) Since the gas is expanding isothermally ( temperature being constant )

work done by the gas

= 2.303 n RT log V₂ / V₁

Here V₂ / V₁ = 4 , T = 298

Put these values in the equation above ,

work done = .075x 2.303 x 8.312 x 298 log 4

= 257.6 J

b) In adibatic change

$pv^\gamma = constant$

$T V^{\gamma-1} = constant$

T₁ / T₂ = $(\frac{V_2}{V_1} )^{\gamma-1}$

298 / T₂ = $( 4)^{1.667-1}$

T₂ = 751.26 K.

In adiabatic change work done

= n R ( T₁ - T₂) / (γ -1)

.075x 8.312 X ( 751 - 298 ) / .667

= 423.38. J