Question

A balloon contains 1.21 x 105 L of ideal gas at 265K. The gas is then cooled to 201 K. What is the volume (L) assuming no gas enters or exits the balloon

Answer 1

The new volume will be 0.918 x 10^5 L

Step-by-step explanation:

initial volume
`V_(1)` = 1.21 x 10^5 L

Initial temperature
`T_(1)` = 265 K

Final volume
`V_(2)` = ?

Final temperature
`T_(2)` = 201 K

Th gas is an ideal gas.

For ideal gases, the equation
`V_(1)`/
`T_(1)` =
`V_(2)`/
`T_(2)` = constant

substituting value, we have

(1.21 x 10^5)/265 =
`V_(2)`/201

`V_(2)` = 24321000/265 = 91777.4 L

= 0.918 x 10^5 L