Aballoon containing 0.500 mol at 0.00°c and 65.0kpa pressure is expanded by adding more argon. how many moles of argon are added to bring the sample to a final volume of 60.0 l at 30.0°c and 45.0kpa

n = 0.572 moles of Argon added to reach a volumen of 60.0 L

Explanation:

∴ n1 = 0.500 mol

∴ T1 = 0.0°C = 273.15 K

∴ P1 = 65.0 KPa * ( 0.009869 atm / KPa ) = 0.6415 atm

∴ R = 0.082 atm.L/K.mol

⇒ V1 = RT1n1 / P1 = 17.46 L

∴ P2 = 45.0 KPa = 0.444 atm

∴ V2 = 60.0 L

∴ T2 = 30.0 °C = 303.15 K

⇒ n2 = ((0.444 atm) * (60.0 L)) / ((0.082 atm.L/K.mol)*(303.15 K))

⇒ n2 = 1.072 mol

moles of Argon added:

n = n2 - n1 = 1.072 - 0.500 = 0.572 mol added

c is the answer because when something is left after something evaporates that shows the chemical change.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml