Answer:
90.6 mL.
Explanation:
Recall the ideal gas law:
[tex]\displaystyle PV = nRT[/tex]
Where P is the pressure; V is the volume; n is the number of moles of gas; R is the universal gas constant; and T is the absolute temperature.
Because only P, V, and T changes, we can rearrange the equation to:
[tex]\displaystyle \frac{PV}{T}= nR[/tex]
The right-hand side is some constant. In other words:
[tex]\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
The initial measurements are 109 kPa, -12.5 °C, and 80.0 mL.
Convert kPa to atm:
[tex]\displaystyle 109\text{ kPa} \cdot \frac{1\text{ atm}}{101.3 \text{ kPa}} = 1.08\text{ atm}[/tex]
And convert temperature to kelvins:
[tex]\displaystyle T_k = (-12.5) + 273.15 = 260.6\text{ K}[/tex]
STP is given by 1 atm pressure and 0 °C (273.15 K). Hence, substitute:
[tex]\displaystyle \frac{(1.08\text{ atm})(80.0\text{ mL})}{(260.6\text{ K})} = \frac{(1\text{ atm})V_2}{(273.15\text{ K})}[/tex]
Solve for volume:
[tex]\displaystyle V_2 = 90.6\text{ mL}[/tex]
In conclusion, the volume of the gas at STP is 90.6 mL.
Check: Volume is indirectly proportional to pressure and directly proportional to temperature. Because we decreased the pressure (1.08 atm to 1.00 atm) and increased the temperature (260.6 K to 273.15 K), we should expect an overall increase in volume.
