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Alecsey [184]
3 years ago
5

A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine

the volume of the sample of CO2(g) if the temperature and pressure are changed to 336 K and 152.0 kPa.
Chemistry
1 answer:
bagirrra123 [75]3 years ago
4 0

Answer:

The final volume of the sample of gas V_{2} = 0.000151 m^{3}

Explanation:

Initial volume V_{1} = 200 ml = 0.0002 m^{3}

Initial temperature T_{1} = 296 K

Initial pressure P_{1} = 101.3 K pa

Final temperature T_{2} = 336 K

Final pressure P_{2} =  K pa

Relation between P , V & T is given by

P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }

Put all the values in the above equation we get

101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )

V_{2} = 0.000151 m^{3}

This is the final volume of the sample of gas.

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If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
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a. 1810mL

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To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

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Adjusting for significant figures, this gives V_2=1810[mL]

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