Answer to Question #269241 in Molecular Physics | Thermodynamics for kat

Question

A 2-kg mass of oxygen expands at a constant pressure of 172 kPa in a piston-cylinder system from a temperature of 32 C to a final temperature of 18 C. Determine (a) the heat required; (b) the work; (c) the change of enthalpy; (d) the change of internal energy.

Answer 1

A 2-kg mass of oxygen expands at a constant-pressure of 172 kPa in piston-cylinder system from a temperature of 32°C to a final temperature of 182°C. [ Oxygen Cp = 0.918 kJ/kg · K]

(a.) Compute heat required in kJ

(b.) Compute steady-flow work in kJ

(c.) Compute total change in Total enthalpy in kJ

$Q= 272.8 \;kJ \\ W= -77.97 \;kJ \\ ΔH=272.8 \;kJ \\ n= \frac{2000}{32} = 62.5 \;mol \\ P=172 \;kPa \\ T_1 = 305 \;K \\ T_2 = 455 \;K \\ V_1 =\frac{nRT_1}{P} \\ = \frac{62.5 \times 8.314 \times 305}{172 \times 10^3} \\ = 0.9214 \;m^3 \\ \frac{V_2}{V_1} = \frac{T_2}{T_1} \\ \frac{V_2}{0.9214} = \frac{455}{305} \\ V_2 = 1.3746 \;m^3 Work = W = -P_{ext}(V_2- V_1) \\ = -172 \times 10^3(\frac{nRT_2}{P} - \frac{nRT_1}{P}) \\ = -nR(T_2 -T_1) \\ = -62.5 \times 8.314 (455- 305) \\ = -77943.75 \;J \\ ΔE = nc_vΔT \\ = 62.5 \times \frac{5}{2} \times 8.314 \times 150 \\ = 194.86 \;kJ$

According to the first law of thermodynamics.

$ΔE = q+W \\ 194.86 = q -77.94 \;kJ \\ q = 272.8 \;kJ \\ ΔH = nc_pΔT \\ = 62.5 \times \frac{7}{2} \times 8.314 \times 150 \;J \\ = 272.8 \;kJ$