Question

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what volume will the gas occupy?

Answer 1

The gas will occupy a volume of 1.702 liters.

Step-by-step explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

`P\cdot V = n\cdot R_(u)\cdot T` (1)

Where:

`P` - Pressure, measured in kilopascals.

`V` - Volume, measured in liters.

`n` - Molar quantity, measured in moles.

`T` - Temperature, measured in Kelvin.

`R_(u)` - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

`(P_(1)\cdot V_(1))/(T_(1)) = (P_(2)\cdot V_(2))/(T_(2))` (2)

Where:

`P_(1)`,
`P_(2)` - Initial and final pressure, measured in kilopascals.

`V_(1)`,
`V_(2)` - Initial and final volume, measured in liters.

`T_(1)`,
`T_(2)` - Initial and final temperature, measured in Kelvin.

If we know that
`P_(1) = 121\,kPa`,
`P_(2) = 202\,kPa`,
`V_(1) = 2.7\,L`,
`T_(1) = 288\,K` and
`T_(2) = 303\,K`, the final volume of the gas is:

`V_(2) = \left((T_(2))/(T_(1)) \right)\cdot \left((P_(1))/(P_(2)) \right)\cdot V_(1)`

`V_(2) = 1.702\,L`

The gas will occupy a volume of 1.702 liters.