If 0.750 L of argon at 1.50 atm and 177$\degree$C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0$\degree$C are added to a 1.00-L flask and the flask's temperature is adjusted to 25.0$\degree$C, what is the resulting pressure in the flask?

Question

Quizplus · Accepted Answer

Step 1: Convert argon pressure to kPa.

$1.50	ext{ atm} 	imes 101.3	ext{ kPa/atm} = 152	ext{ kPa}$

Step 2: Convert argon temperature to Kelvin.

$177	ext{ °F} = (177-32)	ext{ °C} = 145	ext{ K}$

Step 3: Convert sulfur dioxide pressure to atm.

$95.0	ext{ kPa} 	imes 0.00987	ext{ atm/kPa} = 0.937	ext{ atm}$

Step 4: Convert sulfur dioxide temperature to Kelvin.

$63.0	ext{ °F} = (63-32)	ext{ °C} = 291	ext{ K}$

Step 5: Apply combined gas law to find the final pressure.

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$P_1 = 152	ext{ kPa}, V_1 = 0.750	ext{ L}, T_1 = 145	ext{ K}$ (argon)

$P_2 = ?, V_2 = 1.00	ext{ L}, T_2 = 298	ext{ K}$ (final)

$P_3 = 0.937	ext{ atm}, V_3 = 0.235	ext{ L}, T_3 = 291	ext{ K}$ (sulfur dioxide)

First, let's convert the argon volume to liters:

$V_1 = 750	ext{ mL} \div 1000	ext{ mL/L} = 0.750	ext{ L}$

Now we can plug in all the values:

$\frac{(152	ext{ kPa})(0.750	ext{ L})}{145	ext{ K}} = \frac{(P_2)(1.00	ext{ L})}{298	ext{ K}}$

$P_2 = \frac{(152	ext{ kPa})(0.750	ext{ L})(298	ext{ K})}{(145	ext{ K})(1.00	ext{ L})} = 0.946	ext{ atm}$

Finally, we convert the answer to atm and round to the nearest hundredth:

$0.946	ext{ atm} \approx \boxed{	extbf{(C) }0.946	ext{ atm}}$

If 0750 L of Argon at 1 °\degree° C and 0235 L of Sulfur Dioxide at 95

If 0750 L of Argon at 1 $\degree$ C and 0235 L of Sulfur Dioxide at 95