Question

Three 1.0-L flasks, maintained at 308 K, are connected to each other with stopcocks. Initially the stopcocks are closed. One of the flasks contains 1.9 atm of N2, the second 2.8 g of H2O, and the third, 0.70 g of ethanol, C2H6O. The vapor pressure of H2O at 308 K is 42 mmHgand that of ethanol is 102 mmHg. The stopcocks are then opened and the contents mix freely. What is the Pressure? Express your answer to two significant figures.

Answer 1

Data given : 1. Three 1 litre flask

                  2. Temperature = 308 K

                   Flask 1 = 1.9 atm N₂

                   Flask 2 = 2.8 grams H₂O

                  Flask 3 = 0.70 Ethanol

When all the stopcocks are opened and allowed to mix freely the total volume of the system will become 3 Litre. The pressure of N2 = 1.9 atm corresponds to a volume of 1 Litre.

To incorporate this we will do the following

1. Calculate partial pressure of individual components

2. Calculate number of moles corresponding to individual partial pressure

3. Sum all the individual moles obtained corresponding to individual partial pressure

4. Calculate the total pressure corresponding to Volume = 3 Litre and total moles calculate in 3.

Flask 1: N2 is a non-condensable, Hence the all of N2 will remain in gaseous form. Hence the pressure of N2 given in the problem will be the total pressure of N2

   P_N2 =Partial pressure of N₂ =1.9 atm = 1444 mmHg

Partial pressure of N₂ = 1.9 atm

Temperature = 308 K

We will use ideal gas law to calculate moles corresponding to 1.9 atm pressure

P V = N R T                 ...................(1)

Where

P =pressure

V= volume

N= moles

R = ideal gas constant = 0.0821 litre atm mole^-1 K^-1

T=Temperature

1.8 *1 = N *0.0821 * 308

N= 0.0712 moles

Moles of N2 =0.0712 moles

Flask 2&3

Weight of Water =2.8 grams

Weight of ethanol = 0.7 grams

Molar mass of ethanol = 46 grams /moles

Molar mass of water = 18 grams/mole

Number of moles of water = 2.8/18 = 0.156 moles

Number of moles of ethanol = 0.7/46 = 0.0152 moles

Now water and ethanol are condensables, hence some amount would be in liquid and other in liquid.

To calculate partial partial of water and ethanol in vapour ,we will use Raoult's law

Raoults Law is given

P_a= Y_aP = X_a * P^oa                    .............(2)

Where Pa=YaP = partial pressure of component 'a'

         Xa = mole fraction of component 'a' inliquid phase

       P^oa = Vapour pressure of component 'a'

No let us convert given number of moles to moles fraction

Component	Moles	Mole fraction
Water	0.156	0.9112
Ethanol	0.0152	0.0888
Total	0.1712	1

Vapour pressure of water (P^o_water )= 42 mmHg

Vapour pressure of ethanol (P^o_ethanol ) = 102 mmHg

Now we will apply Raoults law to calculate partial pressure

Pwater = Xwater* P^o_water = 0.9112*42 = 38.2704 mmHg

Pethanol = Xethanol* P^o_ethanol = 0.0888*102 = 9.0576 mmHg

Partial pressure of Water = 38.27 mmHg

Partial pressure of ethanol = 9.05676 mmHg

Now let us convert partial pressure to moles by ideal gas law ( equation (1))

For water

P = 38.27 mmHg =38.27/760 mmHg = 0.0504 atm

V= 1 Litre

R =ideal gas constant =0.0821 Litre atm mole^-1 K^-1

T= 308 Kelvin

Substituting in equation (1)

0.0504 *1 = N *0.0821*308

N= 0.001993 moles of water

Moles of water in vapour of phase = 0.001993 moles

For ethanol ,

Pressure = 9.0576 mmHg = 9.0576/760 = 0.01191 atm

Volume = 1 Litre

R =ideal gas constant =0.0821 Litre atm K^-1 Mole^-1

Temperature =308 K

Substituting all the above values in ideal gas equation (1)

0.01191 * 1 = N * 0.0821*308

N =4.7*10^-4 moles of ethanol

Moles of ethanol = 4.7*10^-4 moles

Hence,

Total moles = moles of nitrogen + moles of water + moles of ethanol

                  = 0.0712 + 0.001993 + 4.7*10^-4

                   = 0.073663 moles

Now we will use ideal gas law to calculate pressure corresponding to this total number of moles

N = 0.073663 moles

Volume = 3 Litres

R = ideal gas constant = Litre atm K^-1 Mole^-1

T= 308 K

Substituting all the above values in equation (1)

P * 3 = 0.073663 *0.0821* 308

P =0.6209 atm =0.6209 *760 mmHg = 741.88 mmHg

Pressure is 741.88 mmHg