)An Aeresol can with a volume of 325 mL contains gas at 445kPa at 12ºC. What volume would the gas occupy if it was allowed to escape at 101 kPa and 21ºC? (Answer is 1.5 x 10^3 mL) No matter what I try I can't get the answer right

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)An Aeresol can with a volume of 325 mL contains gas at 445kPa at 12ºC. What volume would the gas occupy if it was allowed to escape at 101 kPa and 21ºC? (Answer is 1.5 x 10^3 mL) No matter what I try I can't get the answer right

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Answer 1 · 2018-05-23T02:12:14

$V_{2} \approx 1500 m l$ $V_2~~1500ml$

Explanation:

Using the combined gas law...

$\frac{P_{1} \cdot V_{1}}{T_{1}}$ $(P_1*V_1)/T_1$ = $\frac{P_{2} \cdot V_{2}}{T_{2}}$ $(P_2*V_2)/T_2$

Solve for $V_{2}$ $V_2$ ...

$V_{2} = \frac{P_{1} \cdot V_{1} \cdot T_{2}}{T_{1} \cdot P_{2} \cdot}$ $V_2=(P_1*V_1*T_2)/(T_1*P_2*)$

$P_{1} = 445 KPa \times \times \times \times \times \times x P_{2} = 101 K P a$ $P_1=445"KPa"color(white)(xxxxxxxxxxxxx)P_2=101KPa$
$V_{1} = 325 m l \times \times \times \times \times \times \times x V_{2} = ?$ $V_1=325mlcolor(white)(xxxxxxxxxxxxxxx)V_2=?$
$T_{1} = (12 + 273) = 285 K \times \times \times T_{2} = (21 + 273) = 294 K$ $T_1=(12+273)=285K"color(white)(xxxxxx)T_2=(21+273)=294K$

$V_2=(325ml*445cancel(KPa)*294cancel(K))/(101cancel(KPa)*285cancel(K))=1477ml~~1500ml$

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)An Aeresol can with a volume of 325 mL contains gas at 445kPa at 12ºC. What volume would the gas occupy if it was allowed to escape at 101 kPa and 21ºC? (Answer is 1.5 x 10^3 mL) No matter what I try I can't get the answer right

1 Answer

Explanation:

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