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jolli1 [7]
3 years ago
14

A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

al pressure is 525 mmHg?
PLEASE HELP, will mark brainliest!!!

Chemistry
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

p_A=\chi_A\times P_T

p_{O_2} = partial pressure of O_2 = ?

\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}

P_{T} = total pressure of mixture  = 525 mmHg

{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles

{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles

Total moles = 1.94 + 1.35 = 3.29 moles

\chi_{O_2}=\frac{1.94}{3.29}=0.59

p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

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a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

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P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

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V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

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