Answer to Question #274653 in Molecular Physics | Thermodynamics for ben

Question

Three kilograms of neon in a constant volume system. The neon is initially at 550 kPa and 350 K. Its pressure is increased to 2000 kPa by a paddle work with 210 kJ of heat added. Find (a) the final temperature, (b) the change in internal energy, (c) the change in enthalpy, (d) the nonflow work. 

Answer 1

$m = 3 \;kg \\ p_1 = 550 \; kPa \\ p_2 = 2000 \; kPa \\ T_1 = 350 \;K \\ Q_{in} = 210 \;kJ$

(a) The final temperature

$Volume = constant \\ pV = mRT \\ \frac{p}{T} = \frac{mR}{T} = constant \\ \frac{p_1}{T_1} = \frac{p_2}{T_2} \\ T_2 = \frac{p_2T_1}{p_1} \\ T_2 = \frac{2000 \times 350}{550} = 1272.73 \;K$

(b) From table Ideal gas specific heat for neon:

$c_p = 1.0299 \; kJ/kg \;K \\ R = 0.4119 \;kJ/kg \;K \\ c_p-c_v =R \\ 1.0299 -c_v = 0.4119 \\ c_v = 0.618 \;kJ/kg \;K$

Change in internal energy

$ΔU =mc_v(T_2-T_1) \\ ΔU = 3 \times 0.618(1272.73-350) = 1710.74 \; kJ$

(c) Change in enthalpy

dH = ΔU + pdV

dH = ΔU = 1710.74 kJ

(d) The nonflow work $= \int pdV = \int p(V_2-V_1) = 0$

$W_{nonflow} = 0$