Phy 1010 lec (7+8).pptx

Ideal Gas Equation

Boyle's Law

When gas is kept at constant temperature its

pressure is inversely proportional to its volume.

Mathematically, Boyle's law can be stated as

EXAMPLE1

A balloon with a volume of 2.0 L is filled with a gas at 3 atmospheres. If the pressure is reduced to 0.5

atmospheres without a change in temperature, what would be the volume of the balloon?

solution

P _i V _i = P _f V _f V _f = P _i V _i /P _f V _i = 2.0 L

P _i = 3 atm P _f = 0.5 atm

V _f = (2.0 L)(3 atm)/(0.5 atm)

V _f = 6 L/0.5 V _f = 12 L

Charle's Law

When the pressure of the gas kept constant

the volume directly proportional to the

temperature.

Example2 :

In one experiment the initial volume and temperature of the gas is 0.5L, 50 C. Assuming the pressure and moles of gas is

constant, what is the volume of the gas if the temperature is increased to 80 0C? Let T1 and V1 be the initial temperature

and volume and let T2, V2 be the final temperature and volume

.

Then according to Charles law

Gay-Lussac's Law

When the volume of the gas kept constant

the pressure directly proportional to the

temperature.

Example

:

10.0 L of a gas is found to exert 97.0 kPa at 25.0°C.

What would be the required temperature (in Celsius) to change the pressure to standard pressure

?

Solution

:

change 25.0°C to 298.0 K and remember that standard pressure in kPa is 101.325

.

Insert values into the equation

:

The ideal gas law

Where n is the number of moles, R is a constant for a specific gas, which can be determined

experimentally, and T is the absolute temperature in Kelvin

The ideal gas law can be expressed in terms of the total number of molecules N where N = nN_A

Example (3)

An ideal gas occupies a volume of 100cm3 at 20oC and

a pressure of 100Pa. Determine the number of moles of gas in the container.

Solution

PV = nRT

Example (4)

Pure helium gas is admitted into a tank containing a movable piston. The initial volume, pressure and temperature of the gas are 15´10-3m3, 200kPa

and 300K respectively. If the volume is decreased to 12´10-3m3 and the pressure is increased to

350KPa, find the final temperature of the gas.

Example (5):

6.2

liters of an ideal gas are contained at 3.0 atm and 37 °C. How

many moles of this gas are present?

Solution

PV = nRT where

n = number of moles of gas

R = gas constant = 0.08 L atm / mol K T = absolute temperature in Kelvin T = 37 °C + 273

T = 310 K n = PV / RT

Heat and the first law of thermodynamics

Heat

The word of ''heat flow'' is an energy

transfer that take place as a consequence of

temperature difference only.

Unit of Heat

The unit of heat is ''calorie'' which is defined as the amount of heat (energy) required to raise the temperature of 1g of water from 14.5oC to 15.5oC_.

Heat capacity and specific heat

The

heat capacity

is defined as the

amount of heat energy needed to raise

The

specific heat capacity

is defined

as the amount of heat energy needed to

raise 1kg of sample by 1 degree Celsius.

Al 900J/kg.Co

Cu 387J/kg.Co

Ag 129J/kg.Co

Example(6)

A 0.05kg of metal is heated to 200oC and then

dropped into a beaker containing 0.4kg of water initially at 20oC. If the final equilibrium

temperature of the mixed system is 22.4oC find the

specific heat of the metal. What is the total heat transferred to water in cooling the metal?

Solution

Heat lost by the metal = heat gained by water m_x c_x (T_i-T_f) = m_w c_w (T_f-T_i)

(0.05Kg) c_x (200oC-22.4oC) = (0.4kg)(4186J/kg.Co)(22.4oC-20oC)

c_x = 453J/kg.Co

Example (7)

A man fires a silver bullet of mass 2g

with a velocity of 200m/sec into a wall.

What is the temperature change of the

bullet?

The kinetic

energy

40J

Q = m c D_T

where c for silver is 234Jkg.Co

D_{T = Q/mc = 85.5C}o

Latent heat

Solid ---- Liquid melting

Liquid --- gas boiling

The heat or energy required to change the

phase

Latent heat of fusion Lf melting (Solid º> Liquid)

Latent heat of vaporization L_v boiling) Liquid º> gas)

L

>

L

Example (8):

How much heat energy is needed to change 0.50 kg of water at 100°C to steam at 100°C?where ℓv=2.3X106 j/kg

Work and heat in thermodynamic processes

dF = P dA

dF dy = P dA dy

dW = P dV

dV=0 W = 0 ( Isochoric ) Process

Isothermal process : at constant temperature Isobaric process : at constant pressure

The first law of thermodynamics

change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work performed by the system on its surroundings. The law can also be stated: The energy

of an isolated system is constant .

Q-W is always constant.

Q-W is called the change in the internal energy of the system

DU

Q and W depend on the path but Q-W is independent of the path

DU = U_f - U_i = Q - W The first law of thermodynamics

for small changes

Q-W is always constant.

Q-W is called the change in the internal

energy of the system

D

U

Q and W depend on the path but Q-W is

independent of the path

DU = U_f - U_i = Q - W The first law of

thermodynamics for small changes

Special cases

In isolated system there is no heat flow

and work is zero the change in internal

energy is zero,

i.e.

D

U=0

If the process is done on a system taken

through a cycle, the change in the internal

Example (9)

One gram of water occupies a volume of 1cm3 _at

atmospheric pressure. When this amount of water is boiled,

it becomes 1671cm3 _{of steam. Calculate the change in}

internal energy for this process.

Where the latent energy of water is (2.26x106J/kg)

Solution

DU = Q - W

Q = m L_v = (1x10-3kg) x (2.26x106J/kg) = 2260J

W = P (V_f - V_i) = (1.013x105) x [(1671-1)10-6] = 169J

Example (10)

A 1kg bar of copper is heated at atmospheric

pressure(1.013x105_N/m2_{) . If its temperature increases from}

20o_{C to 50}o_{C, (a) find the work done by the copper. (b) What}

quantity of heat is transferred to the copper? (c) What is the increase in internal energy of the copper?

Where

intial volume of copper is (1kg/8.92x103_kg/m3₎

the volumetric expansion coefficient β= 5.1x10-5_Co-1

DV = [5.1x10-5Co-1](50oC-20oC)x(1kg/_8.92x103_kg/m3) DV = 1.7x10-7 m3

W = P DV = (1.013x105N/m2) x (1.7x10-7 m3) = 1.9x10-2J

(b) What quantity of heat is transferred to the copper?

Q = m C DT = (1kg) x (387J/kg.Co_{) x (30C}o_{) = 1.16 x 10}4_J

(c) What is the increase in internal energy of the copper?