A mixture of $N_{2}$ $"N"_2$ , $O_{2}$ $"O"_2$ . and ${CO}_{2}$ $"CO"_2$ has a total pressure of $740 torr$ $"740 torr"$ In this mixture, the partial pressure of $N_{2}$ $"N"_2$ is $120 torr$ $"120 torr"$ and the partial pressure of $O_{2}$ $"O"_2$ is $400 torr$ $"400 torr"$ . What is the partial pressure of ${CO}_{2}$ $"CO"_2$ ?

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Answer 1 · 2018-02-24T09:40:48

$220 torr$ $"220 torr"$

From Dalton's Law of Partial Pressures, the total pressure of a mixture of gases is the sum of the partial pressures of each gas.

Therefore, the partial pressure of ${CO}_{2}$ $"CO"_2$ is

$740 torr - 120 torr - 400 torr = 220 torr$ $"740 torr " - " 120 torr " - " 400 torr = 220 torr"$

Answer 2 · 2018-02-24T09:42:47

$220 torr$ $220"torr"$

According to Dalton's Law of Partial Pressures,

$P_"total"=SigmaP_i$ , or $P_"total"=P_1+P_2+....P_n$

Here, we can write the above as:

$P_"total"=P_(N_2)+P_(CO_2)+P_(O_2)$

Here,

$P_"total"=740"torr"$

$P_(N_2)=120"torr"$

$P_(O_2)=400"torr"$

Inputting, we get:

$740=120+400+P_(CO_2)$

$740=520+P_(CO_2)$

$P_(CO_2)=220"torr"$