Question

A piston cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air at 300K. The air undergoes a constant pressure process to a final temp of 400K. During the process heat is transfered to the air by Q=12kJ. 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 on the air and the work done by the air to displace the piston. Gas constant for air is 0.287 kJ/kg*K

note ideal gas model is pV=mRT, and with k=1.4 the total internal energy change of air canbe calculated by U2-U1=(mR/k-1)(delta T)

Answer 1

From first law of thermodynamics-

$Q=\Delta U +W$

Q: Heat supplied to the system

$\Delta U$ : Change in internal energy of the system

W: Work done by the gas/by the system

Here-

Q= 12 kJ

$\Delta U = U_2- U_1 =\frac{mR}{k-1}\Delta T$

$\Delta U =\frac{0.13*0.287}{1.4-1}(400-300)$

$\Delta U =\frac{0.13*0.287*100}{0.4}$

$\Delta U =9.3275kJ$

So-

$12=9.3275+W$

$W=2.6725kJ$

So, net work done by the system is 2.6725kJ.

$W_{net} = W_{air} + W_{paddle}$

$W_{air}$ : work done by the air to displace the piston

$W_{paddle}$: work done by the paddle on the air

as work done by the gas is taken as positive and work done on the gas is taken as negative.so,$W_{air}$ is positve and $W_{paddle}$ is negative.

$W_{air}=\int PdV$

as P is constant.so, we can take it out of the integration.

$W_{air}=P\int dV$

$W_{air}=P(V_2 -V_1)$

$W_{air}=PV_2-PV_1$

from ideal gas model pV=mRT

$W_{air}=mRT_2-mRT_1$

$W_{air}=mR(T_2-T_1)$

$W_{air}=0.13*0.287*(400-300)$

$W_{air}=3.731kJ$

$W_{net} = W_{air} + W_{paddle}$

$2.6725 = 3.731 + W_{paddle}$

$W_{paddle} =2.6725 -3.731$

$W_{paddle} =-1.0585kJ$