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

T= -99 °C.

Step-by-step 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*(1mol)/(28.01g)=0.107mol`

Next, we solve for the temperature as shown below:

`T=(PV)/(Rn)`

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

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

Regards!