A pocket of gas is discovered in a drilling operation. The temperature of the gas is 498º C and its pressure is 13.8 atm. At the surface, the same gas has a volume of l8.0 L at 29º C and 2.00 atm. How large was the pocket of gas originally?
and
A balloon with a pressure of 2.2 atm and a volume of 2.5 L is released in the mountains of Colorado. As the balloon ascends, the volume of the balloon gets noticeably larger and increases to a volume of 4.0 L. What is the new pressure of the balloon?

The answer to your question is below

Explanation:

1)

Data

Temperature 1 = 498°C

Pressure 1 = 13.8 atm

Volume 1 = ?

Temperature 2 = 29°C

Pressure 2 = 2 atm

Volume 2 = 18 L

Process

1.- Convert temperature to °K

Temperature 1 = 498 + 273 = 771 °K

Temperature 2 = 29 + 273 = 302 °K

2.- Use the combine law of gas

              P1V1/T1 = P2V2/T2

Solve for V1

             V1 = T1P2V2 / P1T2

Substitution

             V1 = (771)(2)(18) / (13.8)(302)

Simplification

             V1 = 27756 / 4167.6

Result

             V1 = 6.66 L

2)

Use Boyle's law to solve this problem

             P1V1 = P2V2

Data

P1 = 2 atm

V1 = 2.5 L

P2 = ?

V2 = 4 L

- Solve for P2

             P2 = P1V1/V2

- Substitution

             P2 = (2)(2.5) / 4

- Simplification

             P2 = 5 / 4

- Result

              P2 = 1.25 atm

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

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.