a 7.85 liter container holds a mixture to gases at 15° C the partial pressure of

Question

a 7.85 liter container holds a mixture to gases at 15° C the partial pressure of gas a and Gatsby respectively are 0.273 atmosphere and 0.667 atmosphere. if 0.120moles of a third gas is added with no change in volume or temperature what will the total pressure become

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Explanation / Answer

we have:

P = sum of individual pressures

= 0.273 atm + 0.667 atm

= 0.940 atm

V = 7.85 L

T = 15.0 oC

= (15.0+273) K

= 288 K

find number of moles using:

P * V = n*R*T

0.94 atm * 7.85 L = n * 0.08206 atm.L/mol.K * 288 K

n = 0.3121 mol

This is number of mol of gas present initially

Now calculate the final pressure after 0.120 mol is added

Now we have:

P1 = 0.940 atm

n1 = 0.3121 mol

n2 = 0.3121 mol + 0.120 mol = 0.4321 mol

we have below equation to be used

P1/n1 = P2/Tn2

0.940 atm / 0.3121 mol = P2 / 0.4321 mol

P2 = 1.30 atm

Answer: 1.30 atm

Feel free to comment below if you have any doubts

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