A sample of Freon-12 ( $C F_{2} C l_{2}$ $CF_2Cl_2$ ) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?

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A sample of Freon-12 ( $C F_{2} C l_{2}$ $CF_2Cl_2$ ) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?

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Answer 1 · 2018-05-23T01:48:35

$V_{S T P} = 35.3 L$ $V_(STP)=35.3L$

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 $V_{2}$ $V_2$ ...

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

$P_{1} = 153.3 KPa \times x P_{2} = 101.325 K P a$ $P_1=153.3"KPa"color(white)(xxx)P_2=101.325KPa$
$V_{1} = 25.5 L \times \times \times V_{2} = ?$ $V_1=25.5Lcolor(white)(xxxxxx)V_2=?$
$T_{1} = 298 K \times \times \times T_{2} = 273 K$ $T_1=298K"color(white)(xxxxxx)T_2=273K$

$V_2=(25.5L*153.3cancel(KPa)*273cancel(K))/(101.325cancel(KPa)*298cancel(K))=35.3L$

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A sample of Freon-12 ( $C F_{2} C l_{2}$ $CF_2Cl_2$ ) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?

1 Answer

Explanation:

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A sample of Freon-12 (CF_2Cl_2CF2Cl2CF_2Cl_2) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?

1 Answer

Explanation:

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A sample of Freon-12 ( $C F_{2} C l_{2}$ $CF_2Cl_2$ ) occupies 25.5 L at 298 K and 153.3 kPa. How do you find its volume at STP?