Respuesta :
Explanation:
In an isotropic process, the equation is as follows.
    [tex]\frac{T_{2}}{T_{1}} = [\frac{P_{2}}{P_{1}}]^{\frac{k - 1}{k}}[/tex]
     [tex]T_{2} = T_{1} \times [\frac{P_{2}}{P_{1}}]^{\frac{k - 1}{k}}[/tex]
As the given data is as follows.
    [tex]T_{1}[/tex] = 500 K,   m = 5 kg
    [tex]P_{1}[/tex] = 5 bar,    [tex]P_{2}[/tex] = 1 bar
Now, putting the given values into the above formula as follows.
      [tex]T_{2} = T_{1} \times [\frac{P_{2}}{P_{1}}]^{\frac{k - 1}{k}}[/tex]
      [tex]T_{2} = 500 K \times [\frac{5 bar}{1 bar}]^{\frac{1.4 - 1}{1.4}}[/tex]
           = 315.7 K
As according to the ideal gas equation, PV = mRT
So, calculate the volume as follows.
           V = [tex]\frac{mRT}{P}[/tex]
             = [tex]\frac{5 kg \times 287 J/kg \times 500 K}{5 \times 10^{5}}[/tex]     (as 1 bar = 10^{5} Pa[/tex])
             =  [tex]1.435 m^{3}[/tex]
Now, we will calculate the value of [tex]m_{2}[/tex] as follows.
        [tex]P_{2}V_{2} = m_{2}RT_{2}[/tex]
        [tex]m_{2} = \frac{P_{2}V_{2}}{RT_{2}}[/tex]
            = [tex]\frac{1 \times 10^{5} \times 1.435 m^{3}}{315.7 K \times 287 J/kg}[/tex]
            = 1.58 kg
Thus, we can conclude that the amount of mass remaining in the tank is 1.58 kg and its temperature is 315.7 K.
