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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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 $300 K$ $"300 K"$ and $125.0 kPa$ $"125.0 kPa"$ will be $0.48 L$ $"0.48 L"$

Explanation:

Use the combined gas law to solve this problem. The equation is $\frac{V_{1} P_{1}}{T_{1}} = \frac{V_{2} P_{2}}{T_{2}}$ $(V_1P_1)/T_1=(V_2P_2)/T_2$

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

Given
$V_{1} = 550.0 mL \times \frac{1 L}{1000 mL} = 0.5500 L$ $V_1=550.0"mL"xx(1"L")/(1000"mL")="0.5500 L"$
$P_{1} = 100 kPa$ $P_1="100 kPa"$
$T_{1} = 273.15 K$ $T_1="273.15 K"$
$P_{2} = 125.0 kPa$ $P_2="125.0 kPa"$
$T_{2} = 27^{o} C +273.15=300 K$ $T_2=27^"o""C""+273.15=300"K"$

Unknown
$V_{2}$ $V_2$

Equation
$\frac{V_{1} P_{1}}{T_{1}} = \frac{V_{2} P_{2}}{T_{2}}$ $(V_1P_1)/T_1=(V_2P_2)/T_2$

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

$V_{2} = \frac{V_{1} P_{1} T_{2}}{T_{1} P_{2}}$ $V_2=(V_1P_1T_2)/(T_1P_2)$

$V_{2} = \frac{(0.5500 L \times 100 kPa \times 300 K)}{(273.15 K \times 125.0 kPa)} = 0.48 L$ $V_2=((0.5500"L"xx100"kPa"xx300"K"))/((273.15"K"xx125.0"kPa"))="0.48 L"$ (rounded to two significant figures.

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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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