1) 1.75g of an unknown gas at 31°C and 1.05 atm is stores in a 2.55-L flask. - W

Question

1) 1.75g of an unknown gas at 31°C and 1.05 atm is stores in a 2.55-L flask.

- What is the density of the gas?

______ g/L

- What is the molar mass of the gas?

_____ g/mol

2) If the absolute temperature of a gas is tripled, what happens to the root- mean- square speed of the molecules? (nothing,the new rms speed is 9 times the original rms speed, the new rms speed is 3 times the original rms speed, the new rms speed is 1.732 times the original rms speed, the new rms speed is 1/3 the orginal rms speed)

3) Calculate the average translational kinetic energy (sometimes just called average kinetic energy)...

- for one mole of gas at 723 K.

Ek= ________ J/mol

- for a single gas molecule at 723 K

Ek= ________ J/molecule

4) If He(g) has an average kinetic ebergy of 8890 J/mol inder certain conditions, what is the root mean square speed of F2(g) molecules under the same conditions?

- ___________ m/s

5) Use the van der Waals equation os state to calculate the pressure of 2.70 mol of H2O at 477 K in a 5.10-L vessel.

- P= _________ atm

- use the ideal gas equation to calculate the oressure under the same conditions.

P= _______ atm

Explanation / Answer

Density = mass/volume= 1.75/2.55 g/L=0.686 g/L

From PV= nRT

Number of moles, n = Mass/Molecular weight

Let , Molecular weight= W

1.05*2.55= (1.75/W)*0.0821*(31+273.15)

2.7= (1.75/W)*24.97

W= 1.75*24.97/2.716. =16.18

3.

Root mean square speed = sqrt(3*R*T/M).

M = Molecular weight

V1= Sqrt(3RT/M)

When temperature is triples,

V2= Sqrt(3RT/M)

V2/V1= Sqrt(3)= 1.732

V2= 1.732V1

3. Kinetic energy = (3/2)nRT

n= number of moles =1, T= 723 K, R = 8.314 joules/mole.K

Kinetic energy= 1.5*723*8.314=9016 J/mole=

1 mole contains 6.023*1023 molecules

6.023*1023 molecules will have 9016 joules

1 molecule will   have 9016/(6.023*1023)=1497*10-23 Joules/molecule

4.

From Kinetic energy = 0.5mV2, V= root mean square speed

8890 =(0.5)*4 g/mole*(1kg/1000gm) *V2

V= 2018.5 m/s

5.

Van der waals equation is written as

(P+n2a/V2)*(V-nb)= nRT, where a, b are vander Waals constants

For water , a= 5.3 atm .L2/mole2 and b=0.03049 L.mol

P= nRT/(V-nb)- n2a/V2 =2.7*0.0821*477/(5.1-2.7*0.03049)- 2.7*2.7*5.3/5.1*5.1=19.58 atm

For ideal gas , P= nRT/V= 2.7*0.0821*477/5.1 =20.73 atm

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