A gas occupies 12.3 liters at a temperature of 40.0k what is the volume when the temperature is increased to 60.0k

18.45L

Explanation:

V1 = 12.3L

T1 = 40K

T2 = 60K

V2 = ?

This question involves the use of Charles law which states that the volume of a fixed mass of gas is directly proportional to its temperature provided that pressure remains constant.

Mathematically,

V = kT k = V/T

V1 / T1 = V2 / T2 = V3 / T3 ==Vn / Tn

V1 / T1 = V2 / T2

V2 = (V1 × T2) / T1

V2 = (12.3 × 60) / 40

V2 = 18.45L

The final volume of the gas is 18.45L

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

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