Question

1) If 1.79 g of Ar are added to 2.61 atm of He in a 2.00 L cylinder at 27.0 °C, what is the total pressure of the resulting gaseous mixture?

2) A certain mass of nitrogen gas occupies a volume of 3.85 L at a pressure of 9.94 atm. At what pressure will the volume of this sample be 6.23 L? Assume constant temperature and ideal behavior.

3) A sample of an ideal gas has a volume of 3.10 L at 15.00 °C and 1.30 atm. What is the volume of the gas at 19.80 °C and 0.993 atm?

4) If 32.5 mol of an ideal gas occupies 84.5 L at 89.00 °C, what is the pressure of the gas?

5) How many grams of Kr are in a 3.08 L cylinder at 30.0 °C and 6.64 atm?

Answer #1

*1)*

*Given data ,*

*Temperature = 27 + 273 = 300 K*

*Volume = 2.00 L*

*Weight of Ar = 1.79 gram*

*Molar mass of Ar = 40 gm / mol*

*Moles of Ar = Weight of Ar / Molar mass of Ar*

*= 1.79 / 40*

*= 0.044*

*Let's consider , PV = nRT*

*where P is pressure, T istemperature and V is
volume*

*P = nRT / V*

*= 0.044 *0.0821 * 300 / 2*

*= 1.08 / 2*

*= 0.54 atm*

*The total pressure = (Pressure) _{Argon} +
(Pressure)_{He}*

*= 0.54 + 2.61*

*Total pressure , P = 3.15 atm*

**2)**

**Given data ,**

*Pressure, P1 = 9.94 atm*

*Volume , V1 = 3.85 L*

*Volume , V2 = 6.23 L*

*Let us consider,*

*P1*V1 = P2*V2*

*9.94 atm * 3.85 L = P2 * 6.23 L*

*P2 = 9.94 atm * 3.85 L / 6.23 L*

*= 38.26 / 6.23*

**P2 = 6.14 atm**

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