Question #277242

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 . 


1
Expert's answer
2021-12-10T11:31:14-0500

PV = knT

P = pressure

V = volume

k = constant

m = mass of the gase inside

T = temperature

Pi=6800  kPaVi=0.6  m3mi=25  kgT1=75+273=348  K(6800  kPa)×(0.6  m3)=k×(25  kg)×(348  K)k=0.4689  kPam3/(kgK)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

Vf=Vi=0.6  m3Pf=4130  kPaTf=60+273=333  K(4130  kPa)×(0.3  m3)=(0.4689  kPam3/(kgK))×mf×(333  K)mf=7.93  kgV_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

Vd=Vi=0.6  m3md=mf=7.93  kgTd=75+273=348  KPd×(0.6  m3)=(0.4689  kPam3/(kgK))×(7.93  kg)×(348)Pd=2156.5  kPaV_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


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