Question

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in a 9.0 L flask. The total pressure in the flask was __________ atm. Assume the initial pressure in the flask was 0.00 atm and the temperature upon mixing was 25°C. Select one:

Answer 1

`P=3.7atm`

Step-by-step explanation:

Hello,

In this case, it is possible to determine the pressures of both helium and neon as shown below:

`n_(He)=(P_(He)V_(He))/(RT)=(5.6atm*3.0L)/(0.082(atm*L)/(mol*K)*298.15K) =0.688molHe\\\\n_(Ne)=(P_(Ne)V_(Ne))/(RT)=(3.6atm*4.5L)/(0.082(atm*L)/(mol*K)*298.15K)=0.663molNe`

Now, one considers the total moles (addition between both neon's and helium's moles) and the total volume to compute the final pressure as shown below:

`P=(n_TRT)/(V_T) =((0.688+0.663)mol*0.082(atm*L)/(mol*K)*298.15K)/(9.0L)=3.7atm`

Best regards.