Question #605aa

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Question #605aa

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Answer 1 · 2016-05-03T18:22:06

$V_{2} = 9.06 m L$ $V_2=9.06mL$

Explanation:

Since the number of mole $n$ $n$ is constant and the factors changing are the volume, the pressure and temperature; we can modify the ideal gas law $P V = n R T$ $PV=nRT$ to be:

$\frac{P V}{T} = k$ $(PV)/T=k$ where $k = n R$ $k=nR$ is constant.

Therefore, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

$=>V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)=(0.975cancel(atm)xx8.88mL)/(261cancel(K))xx(273cancel(K))/(1cancel(atm))=9.06mL$

Therefore, the volume of the gas at STP would be $V_2=9.06mL$ .

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Question #605aa

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