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2 years ago

Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?

Chemistry

1 answer:

monitta2 years ago

7 0

Answer:

$200\; \rm mL$ of neon $\rm Ne\, (g)$ at $25^{\circ} \rm C$ and $1\; \rm atm$ (the second choice) would contain an equal number of gas particles as $200\; \rm mL\!$ of $\rm He \, (g)$ at $25^{\circ} \rm C\!$ and $1\; \rm atm\!$ (assuming that all four gas samples behave like ideal gases.)

Explanation:

By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.

All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.

The second choice, $200\; \rm mL$ of neon $\rm Ne\, (g)$ at $25^{\circ} \rm C$ and $1\; \rm atm$ , is the only choice where the volume of the sample is also $200\; \rm mL \!$ . Hence, that choice would be the only one with as many gas particles as $200\; \rm mL\!$ of $\rm He \, (g)$ at $25^{\circ} \rm C\!$ and $1\; \rm atm\!$ .

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