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.
