Question

A sample of gas contains 3.0L of nitrogen at 320kPa. What volume would be necessary to decrease the pressure at 110kPa

Answer 1

`V_(2) = 8.92 L`

Step-by-step explanation:

We have the equation for ideal gas expressed as:

PV=nRT

Being:

P = Pressure

V = Volume

n = molar number

R = Universal gas constant

T = Temperature

From the statement of the problem I infer that we are looking to change the volume and the pressure, maintaining the temperature, so I can calculate the right side of the equation with the data of the initial condition of the gas:

`P_(1) V_(1) =nRT`

`320Kpa*0.003m^(3) =nRT`

`1000L = 1m^(3)`

So

`nRT= 0.96`

Now, as for the final condition:

`P_(2)V_(2)=nRT`

`P_(2) V_(2) =0.96`

clearing
`V_(2)`

`V_(2) =(0.96)/(P_(2) )`

`V_(2) =0.00872m_(3)`

`V_(2) = 8.92 L`