Question

A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = 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 = 480 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:

Change in specific internal Energy$=250\ \rm Btu/lb$

Explanation:

Given:

• Mass of the gas, m=0.4 lb
• Initial pressure and volume are $p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3$

• Final pressure and temperature are $p_1=480\ \rm lbf/in^2$
• Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

$p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3$

So the final volume of the gas is calculated.

Work in any isotropic is given by w

$w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu$

According to the first law of thermodynamics we have

$Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu$

So the Specific Internal Change is given by

$\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb$

So the specific Change in Internal energy is calculated.