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

Answer:

$V_{2} = 8.92 L$

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$V_{2}$

$V_{2} =\frac{0.96}{P_{2} }$

$V_{2} =0.00872m_{3}$

$V_{2} = 8.92 L$