Question

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

Answer 1

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.