Answer to Question #247001 in Molecular Physics | Thermodynamics for Nani

Question

While the pressure remains constant at 689.5 kPa the volume of a system of air changes from 0.567 m3 to 0.283 m³. Determine the heat added/rejected

Answer 1

$P=689.5 \; kPa \\ V_1= 0.567 \;m^3 \\ V_2 = 0.283 \;m^3$

Assume m=1

$Q_{sup} = ΔH = mc_pΔT \\ = 1.005 \;kJ/kgK(T_2-T_1)$

From ideal gas equation

$PV_1 =mRT_1 \\ T_1 = \frac{689.5 \times 0.567}{0.287 \times 1} = 1362.18 \;K \\ PV_2=mRT_2 \\ T_2 = \frac{689.5 \times 0.283}{0.287 \times 1} \\ T_2 = 679.89 \;K \\ Q_{sup} = ΔH = 1.005 \times (679.89-1362.18) = -685.70 \;kJ$