all i really know is that STP (standard temperature and pressure) represents conventional conditions established by IUPAC with T = 273.15 K, P = 1 atm . Now putting these values of temperature and pressure at STP, we can calculate easily volume occupied by 1 mol of gas and which comes out to be 22.4 L. hope this helps
Answer:
Volume of the gas is 90.2 ml
Explanation:
Given:
Initial volume, V1 = 80.0 ml = 0.080 L
Initial pressure, P1 = 109 kPa = 1.076 atm ( 1 kPa = 0.00987 atm)
Initial temperature, T1 = -12.5 C = -12.5 + 273 K = 260.5 K
Under STP conditions i.e. standard temperature and pressure we have:
Pressure, P2 = 1 atm
Temperature, T2 = 273 K
Formula:
Based on the ideal gas relation we have:
[tex]\frac{P1V1}{T1} =\frac{P2V2}{T2} \\\\V2 = \frac{P1V1}{T1} (\frac{T2}{P2} ) = \frac{1.076*0.080}{0.260.5} (\frac{273}{1} )=0.0902 L[/tex]
