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Marina CMI [18]
3 years ago
12

The table below shows a comparison of the different gas laws. Some cells have been left blank.

Chemistry
2 answers:
Nataly [62]3 years ago
7 0
The answer is C.



I hope I helped you


dlinn [17]3 years ago
3 0

Answer 1 :

Name : Boyle's law

Variables : Pressure, volume

Constants : Temperature, moles of gas

Equation : PV = k

Name : Charles’s law

Variables : Volume, temperature

Constants : Pressure, moles of gas

Equation : V = kT

Name : Gay-Lussac's law

Variables : Pressure, temperature

Constants : Volume, moles of gas

Equation : P = kT

Name : Combined gas law

Variables : Pressure, temperature, volume

Constants : Moles of gas

Equation : \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Explanation :

Boyle's Law : It is defined as the pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Charles' Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T

Gay-Lussac's Law : It is defined as the pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

Avogadro's Law : It is defined as the volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

V\propto n

Combined gas law : It is the combination of above four laws.

It is represented as,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Answer 2 : The number of moles are assumed to be constant while using the combined gas law.

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Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

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What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

PV=nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 0.4 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

0.4 liter = 0.4\times10^{-3}=4\times10^{-4} m^{3}

And initial temperature of balloon, T_{1} = 20°C = (273 + 20)K

                                                                          = 293 K

Let the final volume of balloon is V_{2}

And a given, final temperature of balloon, T_{2} is 250°C = (273 + 250)K

                                                                                          = 523 K

Now, V_{1} = KT_{1}

          4\times10^{-4}=K\times293\ (equation\ 1 )

V_{2} = KT_{2}

    =K\times523\ (equation 2)

Dividing equation 1 and 2,

 \frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}

K cancelled by K.

By cross multiplication:

293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\          = \frac{2092\times10^{-4}}{293} \\          =7.14\times10^{-4}m^{3}

Now convert it into liter with the help of calculation done above.

7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

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b) For γ-iron (FCC)

D=2.30*10^{-5} m^{2}/s *exp(-148.0kJ/mol/8.314*10^{-3}kJ/mol*k*1073.15K)= 1.438*10^{-12} m^{2}/s

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