What intermolecular forces are present in each of the substances

What intermolecular forces are present in each of the substances? CH4,C3H8,CH3F,HF, C6H5OH (dispersion forces, dipole-dipole forces, or hydrogen bonding);A sample of ideal gas at room temperature occupies a volume of 34.0L at a pressure of 782torr . If the pressure changes to 3910torr , with no change in the temperature or moles of gas, what is the new volume, V2; and If the volume of the original sample in Part A (P1 = 782torr , V1 = 34.0L ) changes to 65.0L , without a change in the temperature or moles of gas molecules, what is the new pressure, P2. A sample of gas in a balloon has an initial temperature of 9?C and a volume of 1200L . If the temperature changes to 85?C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas? What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 9?C , V1= 1200L ) to a volume of 2400L ? Assume no change in pressure or the amount of gas in the balloon. Oxygen used in respiratory therapy is stored at room temperature under a pressure of 150 atm in a gas cylinder with a volume of 60.0 L.What volume would the oxygen occupy at 0.730atm ? Assume no temperature change; and if the oxygen flow to the patient is adjusted to 6.00L/min at room temperature and 0.730atm , how long will the tank of gas last?

Answer

General guidance

Concepts and reason
Noncovalent interactions: The intermolecular forces that exist among the biomolecules, that are not covalent bonding interactions, are known as noncovalent interactions.

The set of repulsive and attractive forces between biomolecules that results from the polarity is represented as the intermolecular force. There are three types of intermolecular forces:

1)Hydrogen bonding

2)London dispersion force

3)Dipole–dipole force.

The types of intermolecular forces are given below:

Vander Waals forces: These are weak forces of attractions that exist in the molecule.

Dipole-dipole interactions: It occurs between two polar molecules. The interaction between the partial positive charges of one molecule to the partial positive charge of another molecule is represented as dipole–dipole interaction.

Hydrogen bonding: If the hydrogen atom is directly attached to the electronegative atom, it exhibits intermolecular hydrogen bonding.

Ion-ion interactions: The compounds having oppositely charged ions attract each other due to the ion-ion interactions. These are strong intermolecular force of attractions.

Ion-dipole interactions: The interactions present between the ions and dipole of solvent are known as ion-dipole interactions.

London dispersion force: This is a very weak intermolecular force that occurs in the non–polar molecules. The electronegativity difference is very low.

Ideal gas: A gas whose behavior is completely governed by the ideal gas law.

Ideal gas law: Relates the pressure and volume of a given number of moles of gas at a given temperature.

Boyle’s law: It shows the relationship between the pressure and volume of a gas at a constant temperature. Boyle’s law states that the volume of gas held at constant temperature is inversely proportional to its pressure.

Charles’s law: At a fixed amount of gas and constant pressure, the volume is directly proportional to the temperature in Kelvin.

Fundamentals

Ideal gas law:

PV=nRT

Where,

-Pressure of the gas

-Volume of the gas

- Number of moles of gas

-Universal gas constant (8.314JK mol)

-Temperature in Kelvin

Expression for Boyle’s law:

Pav

Where Pressure of the gas and

Volume of the gas

It can also be expressed as follows:

P=kx
PV=K

Where, = Constant.

Therefore, the pressure and volume of two gases at constant temperature could be related as:

P.V=P, V2

Expression for Charles law:

Va T

Where,

-volume of the gas

-temperature

That is,

=a constant,K

For two gases at constant pressure, Charles law can be written as:

V
V2

Step-by-step

Step 1 of 7

(1)

The intermolecular forces present in the given substances are shown below:

CH,
dispersion forces

CH,
dispersion forces

CH,F
dispersion forces
dipole-dipole forces

HF
dispersion forces
dipole-dipole forces
hydrogen bonding

CH,OH
dispersion forces
dipole-dipole forces
hydrogen bonding

Part 1

The intermolecular forces present in the given substances are tabulated below:

CH4
CzHg | CHF
HF
сенон
dispersion forces dispersion forces dispersion forces dispersion forces dispersion forces
dipole - di


Dispersion forces are found in all molecules.

Dipole-dipole forces are present in , and due to polarity.

Hydrogen bonding is present in both and because of strong electronegativity atoms such as fluorine and oxygen.

Step 2 of 7

(2)

The given data in the problem is written below:

P, = 782 torr
V = 34.0L
P, = 3910 torr

The volume of the sample is calculated using the Boyle’s law as:

V
V =

Substitute

78216
V
782 torr x 34.0L
3910 torr
= 6.8L

Part 2

The new volume of the sample is


The initial pressure and volume of the ideal gas in the sample is given to be and respectively. The pressure is increased to at constant room temperature. By substituting these values in the expression for Boyle’s law, the final volume of the sample is .

Step 3 of 7

(3)

The given data in the problem is written below:

P, = 782 torr
V. = 34.0L
V = 65.0L

The pressure of the sample is calculated using the Boyle’s law as:

V
P =

Substitute

P =
782 torr x 34.0L
65.0L
= 409 torr

Part 3

The new pressure of the gas is 409 torr.


The number of moles of an ideal gas and temperature is kept constant. The initial pressure and volume of the ideal gas in the sample is given to be and respectively. The volume is increased to. By substituting these values in the expression for Boyle’s law, the new pressure of the sample is .

Step 4 of 7

(4)

The given data in the problem is written below:

T = 9°C
Vi=1200L
T, = 85 °C

Convert the given temperature into kelvin by using the factor of .

T, =9°C +273
= 282K
T = 85 °C+273
= 358K

The volume of the sample is calculated using the Charles law as:

VT,

Substitute

_1200L x 282K
358K
=945L

Part 4

The new volume of the sample is


The first step in the problem is convert the given temperature into Kelvin by the conversion factor of about 273.

The new volume of the gas is calculated substituting the values of initial volume, initial temperature and final temperature of the gas into the Charles equation.

Step 5 of 7

(5)

The given data in the problem is written below:

T =9°C
V. = 1200L
V2 = 2400L

Convert the given temperature into kelvin by using the factor of .

T, =9°C+273
= 282 K

The temperature of the sample is calculated using the Charles law as:

V.T
T,

Substitute

T,
2400 L 282K
1200L
= 564K

Convert the given temperature into Celsius by using the factor of .

T, = 564K -273
= 291 °C

Hence, the new temperature is 291 °C.

Part 5

The new temperature is291 °C


The new temperature of the gas is calculated substituting the values of initial volume, initial temperature and final volume of the gas into the Charles equation.

Step 6 of 7

(6)

The given data in the problem is written below:

P, = 150 atm
V = 60.0L
P, = 0.730 atm

The volume of the sample is calculated using the Boyle’s law as:

V
V =

Substitute

v_150 atm x 60.0L
0.730 atm
= 12329L

Part 6

The volume of the oxygen gas that occupy at 0.730 atm is 12329 L.


The new volume of the oxygen gas is calculated substituting the values of initial volume, initial pressure and final pressure of the gas into the Boyle’s equation.

Step 7 of 7

(7)

The given data in the problem is written below:

P, = 150 atm
V. = 60.0L
P, = 0.730 atm
rate=6.00 L/min

The calculated volume in the above step is 12329L.

1min
Time required=12329L x-
6.00L
= 2055 min

Part 7

The time needed to completely empty the tank is 2055 min.


The time needed to completely empty the tank is determined by dividing with rate of flow. It is found to be .

Answer

Part 1

The intermolecular forces present in the given substances are tabulated below:

CH4
CzHg | CHF
HF
сенон
dispersion forces dispersion forces dispersion forces dispersion forces dispersion forces
dipole - di

Part 2

The new volume of the sample is

Part 3

The new pressure of the gas is 409 torr.

Part 4

The new volume of the sample is

Part 5

The new temperature is291 °C

Part 6

The volume of the oxygen gas that occupy at 0.730 atm is 12329 L.

Part 7

The time needed to completely empty the tank is 2055 min.

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