Answer to Question #246999 in Molecular Physics | Thermodynamics for Nani

Question

A closed rigid container has a volume of 1 m³ and holds air at 344.8 kPa and 273 K. Heat is added until the temperature is 600 K. Determine the change of internal energy

Answer 1

$V_1= 1 \;m^3 \\ p_1 = 344.8 \;kpa \\ T_1 = 273 \;K \\ T_2 = 600 \;K \\ p_1V_1 =mRT_1 \\ R= 0.287 \;kJ/kg\;K \\ 344.8 \times 1 = m \times 0.287 \times 273 \\ m = 4.4007 \;kg$

The specific heat at constant volume:

$c_v = 0.718 \;kJ/kg \;K$

Internal energy \:

$ΔU = mc_vΔT \\ = 4.4007 \times 0.718 \times (T_2-T_1) \\ = 3.1597 \times (600-273) \\ = 3.1597 \times 327 \\ = 1033.2 \;kJ$

Answer: 1033 kJ