Answer to Question #173333 in Mechanical Engineering for Ade

Question

A container of 5 m ³ contains air at 100 kPa and temperature of 300K. To reduce the pressure and the temperature in the container, part of the air is removed. The new temperature and pressure is recorded as 7 ^oC and 50 kPa respectively. Take R = 287 J/kg.K for air. Obtain the amount of air removed from the container

Answer 1

$V_1 = 5m³\\ P_1 = 100kPa\\ T_1 = 300K$

$T_2 = 7°C + 273K = 280K\\ P_2 = 50kPa\\ V_2 = ?$

$R = 287\ J/kg.K$

$\dfrac{5 × 100}{300} = \dfrac{50 × V_2 }{280}$

$V_2 = 7.33m³$

$∆V = V_2 - V_2 = 7.33 - 5 = 2.33m³$.