A certain gas occupies a volume of 550.0 mL at STP. What would its volume be at 27°C and 125.0 kPa?

1 Answer
Nov 17, 2015

Answer:

The volume of the gas at #"300 K"# and #"125.0 kPa"# will be #"0.48 L"#

Explanation:

Use the combined gas law to solve this problem. The equation is #(V_1P_1)/T_1=(V_2P_2)/T_2#

#"STP"="273.15 K and 100 kPa"#

Given
#V_1=550.0"mL"xx(1"L")/(1000"mL")="0.5500 L"#
#P_1="100 kPa"#
#T_1="273.15 K"#
#P_2="125.0 kPa"#
#T_2=27^"o""C""+273.15=300"K"#

Unknown
#V_2#

Equation
#(V_1P_1)/T_1=(V_2P_2)/T_2#

Solution
Rearrange the equation to isolate #V_2# and solve.

#V_2=(V_1P_1T_2)/(T_1P_2)#

#V_2=((0.5500"L"xx100"kPa"xx300"K"))/((273.15"K"xx125.0"kPa"))="0.48 L"# (rounded to two significant figures.