A certain gas occupies a volume of 550.0 mL at STP. What would its volume be at 27°C and 125.0 kPa?

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Answer 1 · 2015-11-17T03:12:48

The volume of the gas at $"300 K"$ and $"125.0 kPa"$ will be $"0.48 L"$

Use the combined gas law to solve this problem. The equation is $(V_1P_1)/T_1=(V_2P_2)/T_2$

$"STP"="273.15 K and 100 kPa"$

Given
$V_1=550.0"mL"xx(1"L")/(1000"mL")="0.5500 L"$
$P_1="100 kPa"$
$T_1="273.15 K"$
$P_2="125.0 kPa"$
$T_2=27^"o""C""+273.15=300"K"$

Unknown
$V_2$

Equation
$(V_1P_1)/T_1=(V_2P_2)/T_2$

Solution
Rearrange the equation to isolate $V_2$ and solve.

$V_2=(V_1P_1T_2)/(T_1P_2)$

$V_2=((0.5500"L"xx100"kPa"xx300"K"))/((273.15"K"xx125.0"kPa"))="0.48 L"$ (rounded to two significant figures.