Question #b6669

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Question #b6669

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Answer 1 · 2015-04-30T01:07:32

The new volume of hydrogen gas at STP is $4550 L$ $4550 "L"$ .

The equation for the combined gas law is $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ .

If I read your problem correctly, you want the volume of hydrogen gas after it has been adjusted to STP.

Known:
$P_{1} = 100.0 atm$ $P_1=100.0 "atm"$
$V_{1} = 50.0 amt$ $V_1=50.0 "amt"$
$T_{1} = {27.0}^{o} C+273.15=300.2K$ $T_1=27.0^"o""C+273.15=300.2K"$

$P_{2} = 1.00 atm$ $P_2=1.00 "atm"$
$T_{2} = 0^{o} C+273.15=273.15K$ $T_2=0^"o""C+273.15=273.15K"$

Unknown:
$V_{2}$ $V_2$

Solution: Rearrange the equation to isolate and solve for $V_{2}$ $V_2$ .

$V_2=(P_1V_1T_2)/(T_1P_2)=(100.0cancel"atm" * 50.0 "L" * 273.15cancel"K")/(300.2cancel"K" * 1.00cancel"atm")=4550 "L"$

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Question #b6669

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