Respuesta :
Answer:
Explanation:
Given
Pressure, Temperature, Volume of gases is
[tex]P_1, V_1, T_1 & P_2, V_2, T_2 [/tex]
Let P & T be the final Pressure and Temperature
as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas
[tex]-Q=W+U_1----1[/tex]
[tex]Q=-W+U_2-----2[/tex]
where Q=heat loss or gain (- heat loss,+heat gain)
W=work done by gas
[tex]U_1 & U_2[/tex] change in internal Energy of gas
Thus from 1 & 2 we can say that
[tex]U_1+U_2=0[/tex]
[tex]n_1c_v(T-T_1)+n_2c_v(T-T_2)=0[/tex]
[tex]T(n_1+n_2)=n_1T_1+n_2T_2[/tex]
[tex]T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}[/tex]
where [tex]n_1=\frac{P_1V_1}{RT_1}[/tex]
[tex]n_2=\frac{P_2V_2}{RT_2}[/tex]
[tex]T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}[/tex]
[tex]T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}[/tex]
and [tex]P=\frac{P_1V_1+P_2V_2}{V_1+V_2}[/tex]
