Answer:
a) [tex]V_{T} = 9\,m^{2}[/tex], b) [tex]m_{T} = 15\,kg[/tex], c) [tex]P_{T} = 416.667\,kPa[/tex]
Explanation:
a) The equation of state for ideal gas is:
[tex]P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T[/tex]
Given the existence of an isothermal process, the following relation is derived:
[tex]\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}[/tex]
The volume of the other tank is:
[tex]V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}[/tex]
[tex]V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})[/tex]
[tex]V_{2} = 4\,m^{3}[/tex]
The total volume is:
[tex]V_{T} = V_{1} + V_{2}[/tex]
[tex]V_{T} = 5\,m^{3} + 4\,m^{3}[/tex]
[tex]V_{T} = 9\,m^{2}[/tex]
b) The total mass is:
[tex]m_{T} = m_{1} + m_{2}[/tex]
[tex]m_{T} = 5\,kg + 10\,kg[/tex]
[tex]m_{T} = 15\,kg[/tex]
c) The pressure of the gas in the two tanks is:
[tex]P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}[/tex]
[tex]P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)[/tex]
[tex]P_{T} = 416.667\,kPa[/tex]
