1) An idela gas in a sealed container has a initial volume of 2.45 L. At constan

Question

1) An idela gas in a sealed container has a initial volume of 2.45 L. At constant pressure, it is cooled to 19.00°C where its final volume is 1.75 L. Wht was the initial temperature?

- ______°C

2) A sealed container holding 0.0255 L of an idela gas at 0.989 atm and 67°C is placed into a refrigerator and colled to 39°C with no change in volume. Calculate the final pressure of the gas.

- _________atm

3) A sample of an ideal gas has a volume of 3.10 L at 10.20°C and 1.80 atm. What is the volume of the gas at 24.80°C and 0.997 atm?

- ________L

4) If 52.5 mol of an idela gas occupies 63.5 L at 83.00°C, what is the pressure of the gas?

- _____ atm

5) If 2.73 moles of an ideal gas has a pressure of 3.83 atm, and a volume of 26.73 L, what is the temperature of the sample in degrees Celsius?

- ________°C

6) An open flask sitting in a lab fridge looks empty, but we know that actually it is filled with a mixture of gases called air. If the flask volume is 4.00L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?

- ______ molecules

7) A gaseous mixture contains 448.0 Torr of H2(g), 374.5 Torr of N2(g), and 89.7 Torr of Ar(g). Calculate the mole fraction, X, of each of these gases.

- XH2= ________

-XN2=__________

-XAr=____________

Explanation / Answer

1) Ideal gas equation is

P1V1/T1= P2V2/T2

Since P remains constant

V1/T1= V2/T2

V1= 2.45L, T1= 19deg.c= 19+273.15=292.15K

V2= 1.75L, T2= V2*T1/V1= 1.75*292.15/2.45=208.7 K

2)

Since Volume remains constant

P1/T1= P2/T2

P2= P1T2/T1=0.989*(39+273.15)/(67+273.15) =0.907 atm

3.

From P1V1/T1= P2V2/T2

1.8*3.1/(10.2+273.15)= 0.997*V2/ 24.8+273.15)

V2= 5.885 L

4.From PV= nRT, P= nRT/V= 52.5*0.0821*(83+273.15)/63.5=24.17 atm

5.From PV= nRT, T= PV/nR= 3.83*26.73/(2.73*0.0821) K=456.7K=183.55 deg.c

6.

From PV= nRT

Number of moles, n= PV/RT= 1*4/(0.0821*273.15)=0.18 moles

1mole of any gas contains 6.023*1023 molecules

0.18 moles will have 0.18*6.023*1023 molecules=1.08*1023 molecules

7.

Total pressure= sum of partial pressures of the gase- 448+374.5+89.7=912.2 Torr

Partial pressure = Mole fraction * total pressure

Mole fraction = partial pressure/ Total pressure

Mole fractions

H2= 448/912.2=0.46, N2= 374.5/912.2=0.41, Ar= 89.7/912.2= 0.13

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