Question

1. List the types and amounts (volume %) of the four most abundant gases in the atmosphere, excluding water. Use these percentages to calculate the partial pressure in atmospheres of each gas.

a). An aerosol can holds 0.35 L at 4.5 atm pressure. What is the volume when the gas is released into 1.0 atm pressure?

b). Mt. Whitney, in Claifornia, is the highest point in the lower 48 states and is within 150 miles of the lowest place in North America. There is a very popular trail to the summit. However, in order to climb Mt. Whitney you need a permit. The normal pressure at the summit is 440 mm Hg. What is the pressure in atmospheres?

c) At 350 oC a gas occupies 22.4 L. What is the new volume at 700 oC?

d). A weather ballon is filled to 125 L when the atmospheric pressure is 748 mm Hg and the air temperature is 28 oC. It rises to where the atmospheric pressure is 487 mm Hg and the temperature is -15 oC. What is the new volume?

e). What is the quantity (moles) of Neon gas if 4.87 L is at a pressure of 0.635 atm and the temperature is 23 oC?

f). In an experiment, H2 gas was captured over water. The atmospheric pressure was 752.3 mm Hg (equals total pressure of the gas in container). If the water vapor pressure in the container was 24.9 mm Hg, what is the pressure of the hydrogen (only water and hydrogen contribute to the pressure in the container)?

g). Define the heat of fusion. h). Describe what atmospheric pressure is, what it is caused by, and what units it is measured in.

Answer 1

Solution :-

Q1) 4Abundent gases in the atmosphere are as follows

78 % N2 , 21 % O2 , 0.93 % Ar and 0.039 % CO2

Total atmospheric pressure = 1 atm

So lets calculate the partial pressure of the each gas

Partial pressure of the N2 = 1 atm * 78 % / 100 % = 0.78 atm

Partial pressure of the O2 = 1 atm * 21 % / 100 % = 0.21 atm

Partial pressure of the Ar = 1 atm * 0.93 % / 100 % = 0.0093 atm

Partial pressure of the CO2 = 1 atm * 0.039 %/ 100 % =0.00039 atm

a)Aerosol can = 0.35 L and 4.5 atm

volume at 1 atm = ?

P1V1=P2V2

V2 = P1V1/P2

V2= 1.575 L

So the volume of gas released = 1.575 L – 0.35 L = 1.225 L

b) 440 mmHg = ? atm

1 atm = 760 mmHg

Therefore 440 mmHg * 1 atm / 760 mmHg = 0.579 atm

c) given

T1 = 350 C +273 = 623 K

V1= 22.4 L

T2= 700 C + 273 = 973 K

V2= ?

V1/T1 = V2/T2

V2 = V1*T2 / T1

V2 = 22.4 L * 973 K / 623 K

V2 = 35.0 L

So the volume at 700 C is 35.0 L

d) T1 = 28 C +273 = 301 K

P1 = 748 mmHg

V1 = 125 L

T2 = -15 C +273 = 258 K

P2 = 487 mmHg

V2 = ?

Using the combined gas law formula we can calculate the new volume V2 as follows

P1V1/T1 = P2V2/T2

V2 = P1V1*T2 / T1*P2

V2 = 748 mmHg * 125 L *258 K /301 K * 487 mmHg

V2 = 164.6 L

So the new volume is 164.6 L

e) Given data

volume =4.87 L

pressure =0.635 atm

Temperature = 23 C +273 = 296 K

Moles of neon gas = ?

Using the idela gas law formula we can calculate the moles of the neon gas as follows

PV=nRT

n=PV/RT

n= 0.635 atm * 4.87 L / 0.08206 L atm per mol K * 296 K

n= 0.127 mol Neon gas

So the moles of the neon gas present = 0.127 mol Ne     

f) Total pressure =752.3 mmHg

vapor pressure of water = 24.9 mmHg

total pressure = pressure of H2 + vapor pressure of water

therefore

pressure of H2 = total pressure – vapor pressure of water

                           = 752.3 mmHg – 24.9 mmHg

                          = 727.4 mmHg

So the pressure of the H2 gas is 727.4 mmHg

g) Heat of fusion :-

Heat of fusion is defined as the quantity of the ehat needed to change the solid phase into the liquid phase at the constant temperature.

h) Atmospheric pressure :- Atmospheric pressure is the force per unit area exerted against the surface by the weight of the air.

At the sea level 1 square each unit area of the air weighs 14.7 pounds that is lb

So the unit is expressed as 14.7 psi ( pound per square inch)

We can also use the atmospheric unit that 1 atm which is nothing but the 14.7 psi