Question

A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of the same gas at 200 kPa. The two tanks have thermal equilibrium with the ambient environment. After the valve has been opened, the tanks again reach thermal equilibrium with their ambient environment. What is the a) total volume, b) total mass, and c) pressure of the gas in the two tanks

Answer 1

Answer:

a) $V_{T} = 9\,m^{2}$, b) $m_{T} = 15\,kg$, c) $P_{T} = 416.667\,kPa$

Explanation:

a) The equation of state for ideal gas is:

$P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

Given the existence of an isothermal process, the following relation is derived:

$\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}$

The volume of the other tank is:

$V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})$

$V_{2} = 4\,m^{3}$

The total volume is:

$V_{T} = V_{1} + V_{2}$

$V_{T} = 5\,m^{3} + 4\,m^{3}$

$V_{T} = 9\,m^{2}$

b) The total mass is:

$m_{T} = m_{1} + m_{2}$

$m_{T} = 5\,kg + 10\,kg$

$m_{T} = 15\,kg$

c) The pressure of the gas in the two tanks is:

$P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}$

$P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)$

$P_{T} = 416.667\,kPa$