Container A holds 757 mL of ideal gas at 2.60 atm. Container B holds 164 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?

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Container A holds 757 mL of ideal gas at 2.60 atm. Container B holds 164 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?

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Answer 1 · 2016-11-05T03:38:25

The ideal gas equation of state is

$P V = n R T$ $PV = nRT$

In this problem we may consider the temperature to be constant, so we can set up an equation that represents the addition of the number of moles of gas in each container (multiplied through by RT).

$(P_{A}) (V_{A}) + (P_{B}) (V_{B}) = (P_{f}) (V_{f})$ $(P_A)(V_A) + (P_B)(V_B) = (P_f)(V_f)$

$(2.6) (757) + (4.8) (164) = (P_{f}) (757 + 164)$ $(2.6)(757) + (4.8)(164) = (P_f)(757+164)$

$P_{f} = 2.99 a t m$ $P_f = 2.99 atm$

As a check we could assume that $T = 300 K$ $T=300K$ and calculate the moles explicitly using $n = \frac{P V}{R T}$ $n=(PV)/(RT)$ :
$n_{A} = 0.0800 m o l$ $n_A = 0.0800 mol$
$n_{B} = 0.0320 m o l$ $n_B = 0.0320 mol$
$P_{f} = \frac{n R T}{V} = \frac{0.112 \times 0.082054 \times 300}{0.921} = 2.99 a t m$ $P_f = (nRT)/V = (0.112 times 0.082054 times 300)/0.921 = 2.99 atm$

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Container A holds 757 mL of ideal gas at 2.60 atm. Container B holds 164 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?

1 Answer

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