Question

A sample of nitrogen gas in a 750.0 mL sealed flask exerts a pressure of 206.58 kPa. Calculate the temperature of the gas if the flask contains 3.00 grams of gas.

Answer 1

Answer:

T= -99 °C.

Explanation:

Hello there!

In this case, according to the given description of the problem, it is possible for us to analyze this problem via the ideal gas equation:

$PV=nRT$

However, we first calculate the moles in 3.00 grams of nitrogen gas (28.01 g/mol):

$n=3.00g*\frac{1mol}{28.01g}=0.107mol$

Next, we solve for the temperature as shown below:

$T=\frac{PV}{Rn}$

Next, we convert the given pressure and volume to atm and L to obtain:

$T=\frac{(206.58atm/101.325)*750.0L/1000}{0.08206\frac{atm*L}{mol*K}*0.107mol}\\\\T=174.15K-273.15\\\\T=-99\°C$

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