A rigid sealed $3.70 L$ $"3.70 L"$ vessel contains $0.12 mol$ $"0.12 mol"$ of $H e (g)$ $He(g)$ and $0.24 mol$ $"0.24 mol"$ $N e (g)$ $Ne(g)$ at $25^{\circ} C$ $25^@ "C"$ . What is the total pressure in the vessel?

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A rigid sealed $3.70 L$ $"3.70 L"$ vessel contains $0.12 mol$ $"0.12 mol"$ of $H e (g)$ $He(g)$ and $0.24 mol$ $"0.24 mol"$ $N e (g)$ $Ne(g)$ at $25^{\circ} C$ $25^@ "C"$ . What is the total pressure in the vessel?

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Answer 1 · 2016-12-29T02:22:46

"Rigid" containers have constant volume, and we were given that $T = 298.15 K$ $T = "298.15 K"$ . Clearly, the $mol$ $"mol"$ s of gas won't change in this sealed container that houses two (inert) noble gases.

So, we can use the ideal gas law for total volume, total pressure, and total $mol$ $"mol"$ s, assuming ideality of course, to get:

$P V = n R T$ $bb(PV = nRT)$

$\Rightarrow P = \frac{n R T}{V} = \frac{(n_{1} + n_{2}) R T}{V}$ $=> color(blue)(P) = (nRT)/V = ((n_1 + n_2)RT)/V$

$= \frac{(0.12 + 0.24 mols) (0.082057 L \cdot atm/mol \cdot K) (298.15 K)}{3.70 L}$ $= (("0.12 + 0.24 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")$

$=$ $=$ $2.38 atm$ $color(blue)("2.38 atm")$

Another way to do it is via Dalton's Law of partial pressures for an ideal two-gas mixture:

$P = P_{1} + P_{2} + . . .$ $bb(P = P_1 + P_2 + . . .)$

To find $P_{1}$ $P_1$ and $P_{2}$ $P_2$ :

$P_{1} V = n_{1} R T$ $P_1V = n_1RT$

$P_{1} = \frac{n_{1} R T}{V} = \frac{(0.12 mols) (0.082057 L \cdot atm/mol \cdot K) (298.15 K)}{3.70 L}$ $P_1 = (n_1RT)/V = (("0.12 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")$

$=$ $=$ $0.79 atm$ $"0.79 atm"$

Similarly:

$P_{2} = \frac{n_{2} R T}{V} = \frac{(0.24 mols) (0.082057 L \cdot atm/mol \cdot K) (298.15 K)}{3.70 L}$ $P_2 = (n_2RT)/V = (("0.24 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")$

$=$ $=$ $1.59 atm$ $"1.59 atm"$

So that the sum is, again:

$P = P_{1} + P_{2}$ $color(blue)(P) = P_1 + P_2$

$= \frac{n_{1} R T}{V} + \frac{n_{2} R T}{V}$ $= (n_1RT)/V + (n_2RT)/V$

$= 0.79 + 1.59 atm = 2.38 atm$ $= "0.79 + 1.59 atm" = color(blue)("2.38 atm")$

(In a way, we implied Dalton's Law of partial pressures in the first method.)

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A rigid sealed $3.70 L$ $"3.70 L"$ vessel contains $0.12 mol$ $"0.12 mol"$ of $H e (g)$ $He(g)$ and $0.24 mol$ $"0.24 mol"$ $N e (g)$ $Ne(g)$ at $25^{\circ} C$ $25^@ "C"$ . What is the total pressure in the vessel?

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A rigid sealed $3.70 L$ $"3.70 L"$ vessel contains $0.12 mol$ $"0.12 mol"$ of $H e (g)$ $He(g)$ and $0.24 mol$ $"0.24 mol"$ $N e (g)$ $Ne(g)$ at $25^{\circ} C$ $25^@ "C"$ . What is the total pressure in the vessel?