A 1.50 liter flask at a temperature of 25°C contains a mixture of .158 moles of methane, .09 moles of ethane, and .044 moles of butane. What is the total pressure of the mixture inside the flask?

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A 1.50 liter flask at a temperature of 25°C contains a mixture of .158 moles of methane, .09 moles of ethane, and .044 moles of butane. What is the total pressure of the mixture inside the flask?

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Answer 1 · 2016-10-08T13:15:19

$P = \frac{n R T}{V} ≅ 5 \cdot a t m$ $P=(nRT)/V~=5*atm$

Explanation:

Dalton's law of partial pressures holds that in a gaseous mixture, (i) the partial pressure of any component gas is the same as the pressure it would exert if it ALONE occupied the container; and (ii) that the total pressure is the sum of the individual partial pressures.

Thus $P_{total}$ $P_"total"$ $=$ $=$ $P_{methane} + P_{ethane} + P_{butane}$ $P_"methane"+P_"ethane"+P_"butane"$ , and if we (reasonably) assume ideality, then:

$P_{total}$ $P_"total"$ $=$ $=$ $\frac{n_{total} \times 0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 298 \cdot K}{1.50 \cdot L}$ $(n_"total"xx0.0821*L*atm*K^-1*mol^-1xx298*K)/(1.50*L)$

where $(n_{total} = n_{methane} + n_{ethane} + n_{butane})$ $(n_"total"=n_"methane"+n_"ethane"+n_"butane")$ $=$ $=$

$(0.158 + 0.09 + 0.044) \cdot m o l$ $(0.158+0.09+0.044)*mol$ $=$ $=$ $0.292 \cdot m o l$ $0.292*mol$

And so, $P_{total}$ $P_"total"$ $≅$ $~=$ $5 \cdot a t m$ $5*atm$ . You will be able to make a better estimate.

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A 1.50 liter flask at a temperature of 25°C contains a mixture of .158 moles of methane, .09 moles of ethane, and .044 moles of butane. What is the total pressure of the mixture inside the flask?

1 Answer

Explanation:

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