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

Answer:

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

Answer:

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)