Question

Consider two connected rigid vessels separated by a valve. The first rigid vessel has a volume of 0.370 m^3 and contains H2 at 23.5°C and a pressure of 600.×10^3Pa. The second rigid vessel has a volume of 0.710 m^3 and contains Ar at 35.0°C at a pressure of 300.×10^3Pa. The valve separating the two vessels is opened, and both are cooled to a temperature of 12.0°C. Find final pressure

Answer 1

Answer:

380088.7 Pa

Explanation:

$V_1$ = Initial volume = 0.37 m³

$V_2$ = Final volume = 0.71 m³

$P_1$ = Initial Pressure = $600\times 10^3\ Pa$

$P_2$ = Final Pressure = $300\times 10^3\ Pa$

$T_f$ = Final temperature = 12°C

$T_1$ = 23.5°C

$T_2$ = 35°C

We have the equation

$P_f(V_1+V_2)=(n_1+n_2)RT\\\Rightarrow P_f(V_1+V_2)=(\dfrac{P_1V_1}{T_1}+\dfrac{P_2V_2}{T_2})T_f\\\Rightarrow P_f=(\dfrac{P_1V_1}{T_1}+\dfrac{P_2V_2}{T_2})\dfrac{T_f}{V_1+V_2}\\\Rightarrow P_f=(\dfrac{600\times 10^3\times 0.37}{23.5+273.15}+\dfrac{300\times 10^3\times 0.71}{35+273.15})\times \dfrac{12+273.15}{0.37+0.71}\\\Rightarrow P_f=380088.7\ Pa$

The pressure is 380088.7 Pa