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lyudmila [28]
3 years ago
9

A 4.00L flask containing Ne at 25 C and 6.00 atm is joined by a valve to an 8.00 L flask Ar at 25 C and 2.00 atm.

Chemistry
1 answer:
timama [110]3 years ago
6 0

Answer:

P=3.33atm

Explanation:

Hello!

In this case, since know the volume, temperature and pressure of the initial containers, we can compute the moles of each gas prior to the opening of the valve as shown below:

n_{Ne}=\frac{6atm*4L}{0.08206\frac{atm*L}{mol*K}*298.15K} =0.981molNe\\\\n_{Ar}=\frac{2atm*8L}{0.08206\frac{atm*L}{mol*K}*298.15K} =0.654molAr

Next, we add them up to obtain the total moles:

n_T=0.981mol+0.654mol=1.635mol

Now, the total volume:

V_T=4.00L+8.00L=12.00L

Finally, the total pressure is computed by using the ideal gas equation:

P=\frac{1.635mol*0.08206\frac{atm*L}{mol*K}*298.15K}{12.00L}\\\\P=3.33atm

Best regards!

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