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

The volume of the gas at "300 K"300 K and "125.0 kPa"125.0 kPa will be "0.48 L"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_2V1P1T1=V2P2T2

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

Given
V_1=550.0"mL"xx(1"L")/(1000"mL")="0.5500 L"V1=550.0mL×1L1000mL=0.5500 L
P_1="100 kPa"P1=100 kPa
T_1="273.15 K"T1=273.15 K
P_2="125.0 kPa"P2=125.0 kPa
T_2=27^"o""C""+273.15=300"K"T2=27oC+273.15=300K

Unknown
V_2V2

Equation
(V_1P_1)/T_1=(V_2P_2)/T_2V1P1T1=V2P2T2

Solution
Rearrange the equation to isolate V_2V2 and solve.

V_2=(V_1P_1T_2)/(T_1P_2)V2=V1P1T2T1P2

V_2=((0.5500"L"xx100"kPa"xx300"K"))/((273.15"K"xx125.0"kPa"))="0.48 L"V2=(0.5500L×100kPa×300K)(273.15K×125.0kPa)=0.48 L (rounded to two significant figures.