isoprocesses

1. What are isoprocesses?

Solution:

Isoprocesses are processes in an ideal gas with constant mass, during which one of the state variables remains constant.

A) Isothermal process (T = constant)
Boyle–Mariotte’s law

p_1 V_1 = p_2 V_2, \quad \Delta U = 0 \quad \Rightarrow \quad Q = -W = W'

The heat received by an ideal gas during an isothermal process is used to perform work.

B) Isochoric process (V = constant)
Charles’s law

\frac{p_1}{T_1} = \frac{p_2}{T_2}, \quad W = 0 \quad \Rightarrow \quad \Delta U = Q = n c_v \Delta T

The heat received by an ideal gas in an isochoric process is used to increase its internal energy.

C) Isobaric process (p = constant)
Gay–Lussac’s law

\frac{V_1}{T_1} = \frac{V_2}{T_2}, \quad \Delta U = W + n c_p \Delta T

The heat received by an ideal gas during an isobaric process is used both to change internal energy and to perform work.

D) Adiabatic process (Q = 0) – no heat exchange occurs between the gas and the surroundings. $p, V, T$ may change.

Poisson’s law

p_1 V_1^\kappa = p_2 V_2^\kappa

T_1 V_1^{\kappa-1} = T_2 V_2^{\kappa-1}, \quad \Delta U = W = -W'

\kappa = \frac{c_p}{c_v} > 1 \quad \text{Poisson’s constant for the gas

$c_p$ – specific heat capacity at constant pressure
$c_v$ – specific heat capacity at constant volume

2.What volume will 30 liters of air have if, at the same temperature, we increase its pressure from 72 kPa to 75 kPa?

Solution:

Analysis:

V₁ = 30 l = 30·10^-3 m³, p₁ = 72·10³ Pa, p₂ = 75·10³ Pa, V₂ = ?

After increasing the pressure the air will have a volume of 28.8 liters.

3.The density of helium at pressure 10⁵ Pa is ρ₁ = 0.164 kg·m^-3. What will its density be if we compress it to a pressure of 5·10⁷ Pa? (The mass and temperature of the gas do not change during compression).

Solution:

Analysis:

p₁ = 10⁵ Pa, ρ₁ = 0.164 kg·m^-3 , p₂ = 5·10⁷ Pa, ρ₂ = ?

The density of helium after compression will be 82 kg·m^-3.

4.The density of air under normal conditions is ρ₁ = 1.3 kg·m^-3. What will the density of the air be if we heat it from normal conditions to 30^oC? (The amount of air does not change).

Solution:

Analysis:

The density of the air will be 1.17 kg·m^-3.

5.A gas enclosed in a container with a movable piston is heated at constant pressure from 22^oC to 52^oC. By what percentage did its volume increase?

Solution:

Analysis:

T₁ = 295.15 K, T₂ = 322.15 K, Δ% = ?

The volume of the gas after heating increased by 9.1%.

6. A bulb is filled with nitrogen during manufacturing at a pressure of 50.6 kPa and a temperature of 18^oC. What is the temperature of the nitrogen in the lit bulb if its pressure increases to 118 kPa?

Solution:

Analysis:

p₁ = 50.6 kPa = 50.6·10³ Pa, T₁ = 291.15 K, p₂ = 118 kPa = 118·10³ Pa, T₂ =?

The temperature of the nitrogen in the lit bulb is approximately 679 K.

7.At a Formula 1 race, the temperature of the air in the tires increased from 19^oC to 79^oC. How does the pressure in the tire change if it was originally inflated to 240 kPa? Why is a "formation lap" performed before the start? (The internal volume of the tire does not change)

Solution:

Analysis:

The air pressure in the tire after heating increases to approximately 289 kPa. The formation lap ensures that there is no sudden sharp increase in tire pressure and therefore in tire temperature right at the start of the race.

8.In a cylindrical chamber there is air closed by a movable piston located 50 cm from the bottom of the cylinder. The air pressure is 10⁵ Pa. If the piston is moved 20 cm toward the bottom during adiabatic compression, the air pressure increases to 2.05·10⁵ Pa. Determine the Poisson constant for air!

Solution:

Analysis:

h₁ = 0.5 m, V₁ = 0.5S, h₂ = 0.3 m, V₂ = 0.3S, p₁ = 10⁵ Pa,
p₂ = 2.05·10⁵ Pa, κ = ?

The Poisson constant for air is κ = 1.4.

9.Show that for an adiabatic process T₁·V₁^κ-1 = T₂·V₂^κ-1

Solution:

10. For adiabatic compression the volume of the air was reduced to 1/10 of the original volume. Calculate the pressure and temperature of the air after adiabatic compression. The initial air pressure was 10⁵ Pa, the initial temperature 20^oC. κ (air) = 1.4

Solution:

Analysis:

After adiabatic compression the air pressure increased to 2.512 MPa, the temperature to 463.24^oC.

11.The pressure of a gas in a closed container at a temperature of 11⁰C is 189 kPa. What will be the temperature of this gas if its pressure increases to 1 MPa?

Solution:

izodeje-11

The temperature of the gas will increase to 1500 K.

12.The pressure of a gas at a temperature of 20⁰C is 107 kPa. What will its pressure be if its temperature increases to 150⁰C?

Solution:

izodeje-12

The pressure of the gas will increase to 154.4 kPa.

13.The temperature of oxygen with a given mass increases isobarically from an initial temperature of -20⁰C. At what temperature does the oxygen have 1.5 times the volume it had at the initial temperature?

Solution:

izodeje-13

Oxygen has 1.5 times the volume at a temperature of 379.73 K (which is 106.58⁰C)

14.What process is written in the table:

Solution:

izodeje-14r

It is an isothermal process. Boyle–Mariotte’s law applies Boyle–Mariotte's law

15.The volume of a gas is 25 liters with a pressure of 1 kPa. What will its pressure be if we reduce its volume to 20% of the original volume?

Solution:

izodeje-15

The pressure of the gas will increase to 5 kPa.

16.A cylindrical container 30 cm long is closed with a movable piston. A gas in the container is under a pressure of 0.5 MPa. Determine its pressure if the internal volume of the container increases by moving the piston by 10 cm.

Solution:

izodeje-16

The pressure of the gas decreases to 0.375 MPa.

17.The mean free path of a molecule izodeje-17z is at pressure p1 = 10⁵Pa. What will the mean free path of the molecule be at a pressure of

a.) 1.0 Pa,
b.) 10⁷Pa.

Solution:

izodeje-17

18.Hydrogen H₂ has a temperature of -3⁰C. By adiabatic expansion its volume increases to three times the original volume. What is the resulting temperature of the hydrogen during this adiabatic process?

Solution:

izodeje-18

The resulting temperature of the hydrogen is 172.17 K (t = -100.98⁰C).

19.Argon has, at normal pressure p₁ = 101325 Pa, a volume V₁ = 10 liters. After adiabatic compression its volume changes to V₂ = 4 liters and the pressure to p₂ = 0.468 MPa. Determine the Poisson constant for argon.

Solution:

izodeje-19

The Poisson constant for argon is 1.67.

20.Determine the specific heat capacity of argon at constant volume c_V, if the specific heat capacity of argon at constant pressure is

c_P = 525 J.kg^-1.K^-1.

Solution:

izodeje-20

The specific heat capacity of argon at constant volume is c_V = 317 J.kg^-1.K^-1.

Physics

isoprocesses

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