Question

# 1) If 1.79 g of Ar are added to 2.61 atm of He in a 2.00...

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?

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