A. 14.8
b. 17.5
c. 815
d. 1640
e. - 10.9
1. Helium gas in a balloon occupies 2.5 L at 300.0 K. The
balloon is dipped into liquid nitrogen that is at a temperature of
80.0 K. The volume of the helium at the lower temperature is
2. A sample of neon gas has a volume of 752 mL at 25.0'C. by
The volume at 50.0°C will be
ml, if the pressure is
constant
3. A helium-filled balloon has a volume of 2.75 L at 20.0°C
The volume of the balloon changes to 2.46 L when placed
outside on a cold day. The temperature outside is
4. When 1500 L of air at 5.00'C is injected into a household by
furnace, it comes out at 30.0'C. Assuming the pressure is
constant, the volume of the heated air is_
5. A balloon with a volume of 15.5 L is inflated in a room at b
20.0C and then taken outside where the temperature is 7.0C
The volume of the balloon outside, if the pressure remains
constant, will be
6. The volume of gas in a syringe is 15,0 mL at 23.5°C. The by
volume of the gas at 72.5°C will be ml

Explanation:

1)

Given data:

Initial volume  = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

2)

Given data:

Initial volume  = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1  mL

3)

Given data:

Initial volume  = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂  = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75  L

T₂ = 720.78 L.K /  2.75 L

T₂ = 262.1 K

4)

Given data:

Initial volume  = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89  L

5)

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8  L

6)

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79  mL

mechanics (greek   is that area of science concerned with the behaviour of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment. the scientific discipline has its origins in ancient greece with the writings of aristotle and archimedes[1][2][3] (see history of classical mechanics and timeline of classical mechanics). during the early modern period, scientists such as galileo, kepler, and newton laid the foundation for what is now known as classical mechanics. it is a branch of classical physics that deals with particles that are either at rest or are moving with velocities significantly less than the speed of light. it can also be defined as a branch of science which deals with the motion of and forces on objects. the field is yet less widely understood in terms of quantum theory.

explanation:

b is the correct