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"#?
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#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"#
The final pressure is 1560 mmHg.
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)
