Answer:
The resulting pressurei n the flask is 0.941 atm
Explanation:
Step 1: Data given
Volume of argon = 0.750 L
Pressure argon = 1.50 atm
Temperature = 177 °C = 450 K
Volume of SO2 = 0.235 L
Pressure of SO2 = 713 torr = 713/760 atm = 0.938
Temperature = 63.0 °C = 336 K
gasses are added to a 1.0 L flask with a temperature of 25.0 °C
Step 2:
From the ideal gas law:
p*V = nRT
⇒ the number of moles n = constant
⇒ the gas constant R = constant
so: P1 * V1 / T1 = P2 * V2 / T2.
⇒ The argon and the sulfur dioxide become one gas mixture sharing pressure, volume, and temperature in the flask at the end.
We can write this as:
Pa * Va / Ta + Ps * Vs / Ts = P2 * V2 / T2
⇒ with Pa = the pressure of argon = 1.50 atm
⇒ with Va = the volume of argon = 0.750 L
⇒ with Ta = the temperature of argon = 450.15 K
⇒ with Ps = the pressure of SO2 = 0.938 atm
⇒ with Vs = the volume of SO2 = 0.235 L
⇒ with Ts = the temperature of SO2 = 336.15 K
1.5 atm * 0.750 L / 450.15 K + 0.938 atm * 0.235 L / 336.15 = P2 * 1 L / 298.15 K
P2 = 0.941 atm
The resulting pressure in the flask is 0.941 atm
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