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)

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

T₁ / T₂ =

298 / T₂ =

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

b

explanation:

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever