Apiston–cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air, initially at 300 k. the air undergoes a constant-pressure process to a final temperature of 460 k. during the process, energy is gradually transferred to the air by heat transfer in the amount 12 kj. assuming the ideal gas model with k = 1.4 and negligible changes in kinetic and potential energy for the air, determine the work done by the paddle wheel on the air and by the air to displace the piston, each in kj.

Explanation:

From the first law of thermodynamics

Q=ΔU+W

Q=heat supplies to the system

ΔU=change in internal energy of the system

W= work done by the gas  by the system

Q =12 KJ

ΔU==(mR/k-1)ΔT

ΔU=*(460-400)

     =14.924 KJ

Q=ΔU+W

W=-2.924 KJ

=+

=work done by the air

= P(V2-V1)

=mRT2-mRT1

=mR(T2-T1)=0.138*0.287*(460-300)=

                           =6.33 KJ

=+

                      =-2.924-6.33

=-9.26  KJ                    

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.