Question

A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.1 = constant. The mass of the gas is 0.4 lb and the following data are known: p1 = 160 lbf/in.2, V1 = 1 ft3, and p2 = 300 lbf/in.2 During the process, heat transfer from the gas is 2.1 Btu. Kinetic and potential energy effects are negligible. Determine the change in specific internal energy of the gas, in Btu/lb.

Answer 1

Answer:

$\Delta u = 53.99 Btu/lbm$

Explanation:

given data:

$P_1 = 160 lbf/in^2 = 23040 lbf/ft^2$

$P_2 = 300 lbf/in^2 = 56160 lbf/ft^2$

$V_1 = 1ft^3$

Q = - 2.1 Btu

$PV^{1.1} = constant$

$P_1V_1^{1.1} = P_2V_2^{1.1}$

$V_2 = V_1 [\frac{P_1}{P_2}]^{1.1}$

       $= 1 [\frac{23040}{56160}]^{\frac{1}{1.1}$

       $= 0.44 ft^3$

work done during polytropic process

$w = \frac{P_2V_2 - P_1V_1}{1-n}$

   $= \frac{56160 \times 0.4759 - 23040\times 1}{1 - 1.2}$

   = -18442.1 ft lbf

we know 1 Btu = 778.169 ft. lbf, therefore

w = -23.6993 Btu

$Q = W + m\Delta u$

$-2.1 = - 23.6993 + 0.4 \Delta u$

$\Delta u = 53.99 Btu/lbm$