A tank has a volume of 0.1 m3 and is filled with He gas at a pressure of 5 x 106 Pa. A second tank has a volume of 0.15 m' and is filled with He gas at a pressure of 6 x 106 Pa. A valve connecting the two tanks is opened. Assumıng He to be a monatomic ideal gas and the walls of the tanks to be adiabatic and rigıd, find the final pressure of the system Hint: Note that the internal energy is constant.

Consider that, as the system is adiabatic, [tex]U_{1}= U_{2}[/tex] where U1 and U2 are the internal energies before the process and after that respectively.

Consider that: [tex]U=H-PV[/tex], and that the internal energy of the first state is the sum of the internal energy of each tank.

So, [tex]H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}[/tex]

Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: [tex]H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}[/tex]

So, [tex]P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}[/tex]

Isolating [tex]P_{2}[/tex],

[tex]P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}[/tex]

[tex]V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}[/tex]

So,

[tex]P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa[/tex]