Question

If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 713 torr and 63.0°C are added to a 1.00-L flask and the flask's temperature is adjusted to 25.0°C, what is the resulting pressure in the flask?

Answer 1

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