A student collects 450 ml. of a gas at a pressure of 90.0 kpa and a temperature of 17.0 degrees C. What is the new volume of the gas at 0.0 degrees C and a pressure 110.KPA?

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We can use the combined gas law to solve this problem. It is expressed as:

`(P_1 V_1)/(T_1) = (P_2 V_2)/(T_2)`

- P1 = 90 kPa = 0.8883 atm
- V1 = 450 mL = 0.45 L
- T1 = 17 + 273.15 = 290.15 K

- P2 = 110 kPa = 1.0857 atm
- V2 = ?
- T2 = 0 + 273.15 = 273.15

For pressure, we convert the unit from kPa to atm, degrees Celsius to K and for the volume form mL to L.

1 kPa = 0.00987atm

1 L = 1000 mL

K= degree Celsius + 273.15

Substitute the given values and solve algebraically.

`(0.8883*0.45)/(290.15) = (1.0857*V_2)/(273.15)`

`1.378x10^(-3) = 3.975x10^(-3) *V_2`

`V_2 = (1.378x10^(-3))/(3.975x10^(-3))`

`V_2 = 0.347 L = 347 mL` **-> answer**

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