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Kitty [74]
3 years ago
5

Explain the laws of thermodynamics and why they matter.

Chemistry
1 answer:
creativ13 [48]3 years ago
7 0
Following are the laws of thermodynamics, with suitable example. 

1st Law of thermodynamics: 
1st law of thermodynamics deals with conservation of energy. It is stated as '<span> energy cannot be created or destroyed in an isolated system'. According to this law, total energy of universe remains constant. Energy just gets converted from one form to another. For example, in case of burning of cracker, chemical energy stored in cracker is converted into heat, light and sound energy.

2nd Law of thermodynamics:
2nd law of thermodynamics deals with entropy change associated with system. It is stated as '</span><span> entropy of any isolated system always increases'. According to this law, the system tries to maximize entropy. System with higher entropy is more stable than system with lower entropy. For instance, at room temperature, ice melts into water, because water has higher entropy than ice. It may be noted that entropy is measure of disorder in system. Thus, higher the disorder in system, greater is the entropy.

3rd Law of thermodynamics:
3rd law of thermodynamics also deal with entropy change in system. According to this law, </span><span>entropy of a system approaches a constant value as the temperature approaches absolute zero. This means that, as the temperature decreases, randomness in system decreases and finally at 0K, system is in highly order state, hence ideally system must have zero entropy. However, there is always some residual entropy present in system even at 0K, due structural orientation of molecules.  </span>
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The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO
Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy =  435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435  +  431 + x

x = -92 - (435  +  431)

x = -958 kJ mol^-1

Learn More about dissociation enthalpy:brainly.com/question/9998007?

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6 0
2 years ago
The volume of 160. g of CO initially at 273 K and 1.00 bar increases by a factor of two in different processes. Take CP,m to be
mel-nik [20]

Answer:

Explanation:

w eFedfweF edf SEDFAsFGFSDSFG BAEWRDA G

5 0
3 years ago
In a chemical reaction 88g of CO2 was produced at r.t.p. <br> Calculate the volume of CO2 in dm??
Tems11 [23]

Explanation:

RFM = 44 \: g \\ 44g \: contain \: 1 \: mole \\ 88g \: will \: contain \: ( \frac{88}{44}  \times 1) \\  = 2 \: moles \\ at \: room \: temperature \\ 1 \: mole \: occupies \: 24 \:  {dm}^{3}  \\ 2 \: moles \: occupies \: (2 \times 24) \\  = 48 \:  {dm}^{3}

6 0
3 years ago
At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival
MissTica

Answer:

The correct answer is 1.5 balloons.

Explanation:

Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,  

= 4/3 × π × (170)³ cm³

= 20.5176 × 10⁶ cm³  

= 20.5176 m³

The density of air at 100 degree C s 0.946 Kg m⁻³

The mass of air in each balloon can be calculated by using the formula,  

Mass = density × volume

Mass = 0.946 Kg m⁻³ × 20.5176 m³

Mass = 19.410 Kg

The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,  

= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C

= 14.68 × 10⁵ J

The concentration of butane given is 1.00 Kg or 1000 g

The molecular weight of butane is 58.12 g per mole

The moles or n can be calculated by using the formula,  

n = mass / mol.wt

n = 1000 g/58.12 g/mol = 17.20 mol

The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,  

= 125.7 × 10³ J/mol × 17.20 mol

= 2.16 × 10⁶ J

The number of balloons that can be inflated with hot air is,  

= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon

= 1.5 balloons

Hence, maximum of 1.5 balloons can be inflated.  

7 0
4 years ago
What is the volume of 500g of CO2?
Airida [17]

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter


Try to see if this helps

3 0
3 years ago
Read 2 more answers
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