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"?

2 Answers

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"

Jan 13, 2016

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)