Question

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.

Answer 1

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Step-by-step explanation:

Given that;

`V_(A)` = 1 m³

`T_(A)` = 10°C = 283 K

`P_(A)` = 350 kPa

`m_(B)` = 3 kg

`T_(B)` = 35°C = 308 K

`P_(B)` = 150 kPa

Now, lets apply the ideal gas equation;

`P_(B)`
`V_(B)` =
`m_(B)`R
`T_(B)`

`V_(B)` =
`m_(B)`R
`T_(B)` /
`P_(B)`

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

`V_(B)` = ( 3 × 0.287 × 308) / 150

`V_(B)` = 265.188 / 150

`V_(B)` = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also,
`m_(A)` =
`P_(A)`
`V_(A)` / R
`T_(A)` = (350 × 1)/(0.287 × 283) = 350 / 81.221

`m_(A)` = 4.309 kg

Total mass,
`m_(f)` =
`m_(A)` +
`m_(B)` = 4.309 + 3 = 7.309 kg

Total volume
`V_(f)` =
`V_(A)` +
`V_(B)` = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

`P_(f)` =
`m_(f)`R
`T_(f)` /
`V_(f)`

given that; final temperature
`T_(f)` = 20°C = 293 K

we substitute

`P_(f)` = ( 7.309 × 0.287 × 293) / 2.77

`P_(f)` = 614.6211119 / 2.77

`P_(f)` = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa