Question

What pressure is required to contain 0.644 grams of nitrogen gas (N2) in a 4.2 L container at a temperature of 20 degrees Celsius?

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Answer 1

0.13 atm pressure is required to contain 0.644 grams of nitrogen gas ($N_{2}$) in a 4.2 L container at a temperature of 20 degrees Celsius.

What is an ideal gas equation?

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in $m^3$.

Calculating moles of $N_{2}$:

Moles = $\frac{Given mass}{ molar mass}$

Moles = $\frac{0.644 }{ 28}$

Moles = 0.023

Now, using the Ideal Gas Law;

PV=nRT

We can rearrange to solve for pressure:

P=nRT/V

P is pressure, n is the number of moles (0.023), R is a constant (0.08206 L*atm/mol*K), T is the temperature in Kelvin (293.15K) and V is volume, 4.2 L.

So, plugging in;

P=$\frac{(\;0.023\;mol)(0.08206\;L*\;atm\;/\;mol\;*\;K)(\;293.15\;K)}{(\;4.2\; L)}$

P=0.1317 atm

Hence, 0.13 atm pressure is required to contain 0.644 grams of nitrogen gas ($N_{2}$) in a 4.2 L container at a temperature of 20 degrees Celsius.

Learn more about the ideal gas equation here:

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