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Rufina [12.5K]
3 years ago
11

Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

Chemistry
1 answer:
maria [59]3 years ago
6 0

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also  decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.  Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

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Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
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Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

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