Question

A gas at 20.0°C and 52.1 mL is heated to 93.5°C. What is the new volume?

Answer 1

`( V_(1) )/( T_(1) )` =
`( V_(2) )/( T_(2) )`

`V_(1)` = 0.0521 L

`T_(1)` = 293.15 K

`T_(2)` = 366.65 K

Solve for
`V_(2)`


`(0.0521 L)/(293.15 K)` =
`\frac { V_(2) }{366.65 K}`


`V_(2)` = 0.06516 L or 65.2 mL