Question

A `"2270 mL"` sample of gas has a pressure of `"570. mmHg"` at `"25"^@"C"`. What is the pressure when the volume is decreased to `"1250 mL"` and the temperature is increased to `"175"^@"C"`?

Answer 1

1556 mm Hg

Explanation:

`V_1/V_2= (T_1P_2)/(T_2P_1)`

`T_1 = (25 +273.15) K = 298.15 K; T_2 = (175 +273.15) K = 448.15 K`

`=> P_2 = (V_1T_2P_1)/(V_2T_1)`

`=> P_2 = (2270 color(red)(cancel(color(black)("mL")))*448.15 color(red)(cancel(color(black)("K")))×"570. mm Hg")/ (1250color(red)(cancel(color(black)("mL")))*298.15 color(red)(cancel(color(black)("K"))))`

`=> P_2 = "1556 mm Hg"`

Answer 2

The final pressure is 1560 mmHg.

Explanation:

Use the combined gas law, which relates pressure, volume, and temperature.

`(P_1V_1)/T_1=(P_2V_2)/(T_2)`

Note: Temperature must be converted from degrees Celsius to Kelvins.

Known/Given
`P_1="570. mmHg"`
`V_1="2270 mL"`
`T_1="25"^"o""C"+273.15="298 K"`
`V_2="1250 mL"`
`T_2="175"^"o""C"+273.15="448 K"`

Unknown
`P_2="???"`

Solution
Rearrange the equation to isolate `P_2` and solve.
`(P_1V_1)/T_1=(P_2V_2)/(T_2)`

`P_2=(P_1V_1T_2)/(T_1V_2)`

`P_2=(570."mmHg"xx2270cancel"mL"xx448cancel"K")/(298cancel"K"xx1250cancel"mL")="1560 mmHg"` (rounded to three significant figures)