Question

Nitrogen (N2) contained in a piston–cylinder arrangement, initially at 10 bar and 405 K, undergoes an expansion to a final temperature of 300 K, during which the pressure–volume relationship is pV1.3 = constant. Assuming the ideal gas model for the N2, determine the heat transfer in kJ/kg.

Answer 1

Answer:28.21 kJ/kg

Explanation:

Given

$P_1=10\ bar$

$T_1=405\ K$

$T_2=300\ K$

Process $PV^{1.3}=constant$

Work done for Polytropic process

$W=\dfrac{P_1V_1-P_2V_2}{n-1}$

where n=Polytropic index

$W=\dfrac{R(T_1-T_2)}{n-1}$

$W=\dfrac{0.296(405-300)}{1.3-1}\quad [R_{N_2}=\frac{8.314}{28}]$

$W=103.6\ kJ\kg$

Now Calculating change in Internal energy

$\Delta U=c_v(T_2-T_1)$

$\Delta U=0.718\times (300-405)$

$\Delta U=-75.39\ kJ/kg$

Now applying First law concept

$\Delta U=Q-W$

$Q=W+\Delta U$

$Q=103.6-75.392$

$Q=28.21\ kJ/kg$