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Gas laws from KTG

Which of the ...

Question

Which of the following options is/are correct with respect to $C H_{4}$ and $S O_{2}$ ? (Equal mass of $C H_{4}$ and $S O_{2}$ are contained in a 2 L container at same temperature)

A

C H_{4}

will diffuse 4 times faster than

S O_{2}

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B

Average speed of

C H_{4}

is 2 times that of

S O_{2}

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C

Both have the same average kinetic energy per mol

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D

C H_{4}

has double the partial pressure of

S O_{2}

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Solution

The correct options are
B Average speed of $C H_{4}$ is 2 times that of $S O_{2}$
C Both have the same average kinetic energy per mol
Moles of $C H_{4} =$ 4 times of moles of $S O_{2}$ , So $p_{C H_{4}} = 4 p_{S O_{2}}$

$\frac{r_{C H_{4}}}{r_{S O_{2}}} = \frac{p_{C H_{4}}}{p_{S O_{2}}} \sqrt{\frac{M_{S O_{2}}}{M_{C H_{4}}}} = \frac{4}{1} \sqrt{\frac{64}{16}} = 8$

$\frac{{¯ V}_{C H_{4}}}{{¯ V}_{S O_{2}}} = \sqrt{\frac{M_{S O_{2}}}{M_{C H_{4}}}} = 2$
K.E. (per mol) $= (\frac{3}{2}) R T$ , So same for both.
Partial pressure of a gas is directly proportional to mole fraction of that particular gas.

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Similar questions

Q. Which of the following options is/are correct with respect to

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