Question

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the temperature changed to:

(a) -5 ºC
(b) 95 ºF
(c) 1095 K

Answer 1

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Step-by-step explanation:

Given:
`V_(1)` = 571 mL,
`T_(1) = 26^(o)C`

(a)
`T_(2) = 5^(o)C`

The new volume is calculated as follows.

`(V_(1))/(T_(1)) = (V_(2))/(T_(2))\\(571 mL)/(26^(o)C) = (V_(2))/(5^(o)C)\\V_(2) = 109.81 mL`

(b)
`T_(2) = 95^(o)F`

Convert degree Fahrenheit into degree Cesius as follows.

`(1^(o)F - 32) * (5)/(9) = ^(o)C\\(95^(o)F - 32) * (5)/(9) = 35^(o)C`

The new volume is calculated as follows.

`(V_(1))/(T_(1)) = (V_(2))/(T_(2))\\(571 mL)/(26^(o)C) = (V_(2))/(35^(o)C)\\V_(2) = 768.65 mL`

(c)
`T_(2) = 1095 K = (1095 - 273)^(o)C = 822^(o)C`

The new volume is calculated as follows.

`(V_(1))/(T_(1)) = (V_(2))/(T_(2))\\(571 mL)/(26^(o)C) = (V_(2))/(822^(o)C)\\V_(2) = 18052.38 mL`