Answer to Question #277242 in Molecular Physics | Thermodynamics for ken

Question

A 0.6 m3 tank contains gas at a pressure of 6800 kPa gage, a temperature of 75oC and mass of 25 kg. A part of this gas was discharged and the temperature and pressure changed to 60oC and 4130 kPa respectively. Heat was applied and the temperature was back to 75oC. Find the final mass, discharged mass, and the final pressure of the gas . 

Answer 1

PV = knT

P = pressure

V = volume

k = constant

m = mass of the gase inside

T = temperature

$P_i = 6800 \; kPa \\ V_i = 0.6 \; m^3 \\ m_i = 25 \; kg \\ T_1 = 75 + 273 = 348 \;K \\ (6800 \;kPa) \times (0.6 \; m^3) = k \times (25 \; kg) \times (348 \;K) \\ k = 0.4689 \; kPa \cdot m^3/(kg \cdot K)$

After the discharge of gas

$V_f=V_i = 0.6 \; m^3 \\ P_f = 4130 \;kPa \\ T_f = 60 + 273 = 333 \;K \\ (4130 \;kPa) \times (0.3 \;m^3) = (0.4689 \; kPa \cdot m^3/(kg \cdot K)) \times m_f \times ( 333 \;K) \\ m_f = 7.93 \; kg$

Final mass = 7.93 kg

Discharged mass = 25 – 7.93 = 17.07 kg

After applying the heat

$V_d = V_i = 0.6 \;m^3 \\ m_d=m_f = 7.93 \; kg \\ T_d = 75+273 = 348 \;K \\ P_d \times (0.6 \;m^3) = (0.4689 \; kPa \cdot m^3/(kg \cdot K)) \times (7.93 \; kg) \times (348) \\ P_d = 2156.5 \; kPa$