Question

A rigid tank contains 1 kg of N2 at 25°C and 300 kPa is connected to another rigid tank that contains 3 kg of O2 at 25°C and 500 kPa. The valve connecting the two tanks is opened, and the two gases are allowed to mix. If the final mixture temperature is 25°C.

Required:
Determine the volume of each tank and the final mixture pressure.

Answer 1

Following are the solution to this question:

Step-by-step explanation:

let,

`\text{m= mass of the gas}\\\text{M= molar weight of the gas}\\ m_(N_2) = 1 \kg \\T_(N_2)= 298 \ K\\ M_(N_2) = 28 \ (kg)/(kmol)\\ P_(N_2)= 300 \ kPa\\m_(O_2) = 3 \ kg \\T_(O_2) = 298 \ K \\M_(O_2)=32 \ (kg)/(kmol)\\ P_(O_2) = 500 \ kPa \\T_m= 298 \ K`

Calculating the mole of nitrogen:

`N_(N_2) = (m_(N_2))/(M_(N_2))`

`=(1)/(28) \\\\= 0.0357 \ kmol`

Calculating the mole oxygen:

`N_(O_2) = (m_(O_2))/(M_(O_2))\\\\=(3)/(32) \\\\= 0.09375 \ kmol`

Calculating the total mole:

`N_m=N_(N_2)+N_(O_2)\\\\= 0.0357+0.09375\\\\= 0.1295 \ kmol`

Calculating the volume of nitrogen:

`V_(N_2) =(N_(N_2) * R_(u)* T_(N_2))/(P_(N_2))\\\\= (0.0357 * 8.314 * 298)/(300)\\\\= 0.295 m^3`

Calculating the volume of oxygen:

`V_(O_2) =(N_(O_2) * R_(u)* T_(O_2))/(P_(O_2))\\\\= (0.09375 * 8.314 * 298)/(500)\\\\= 0.465 m^3`

Calculating the total volume:

`V_m=V_(N_2)+V_(O_2)\\\\=0.295+0.465 \\\\=0.76 \ m^3`

Calculating the mixture pressure:

`P_m = (N_(m) * R_u * T_m)/(V_m) \\\\= (0.1295 * 8.314 * 298)/(0.76) \\\\= 422.2\ kPa`