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Question: At \(5 \times {10^5}\) bar pressure, the density of diamond and graphite are \[3g/cc\] and \[2g/cc\]...

At 5×1055 \times {10^5} bar pressure, the density of diamond and graphite are 3g/cc3g/cc and 2g/cc2g/cc respectively. At a certain temperature “T”. find the value of ΔUΔH\Delta U - \Delta H for the conversion of 1 mole of graphite to 1 mole of the diamond at temperature” T”.

A.100KJ/mol

B. 50KJ/mol

C. -100KJ/mol

D. None of these

Explanation

Solution

Enthalpy is a thermodynamic quantity and is often used by scientists to measure the energy changes in the chemical reactions and it is defined as the sum of internal energy of the system (U) and product of pressure (P) and Volume (V).

Complete step by step answer:

Since the diamond and graphite are the allotropes of carbon. Hence the weight of 1mole of graphite and diamond is 12 grams. As we know the relation between change in internal energy and change in enthalpy is given as:

ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V (1)

Here, the change in enthalpy is represented by ΔH\Delta H and the change in internal energy is represented by ΔU\Delta U and the Pressure is represented by P. The change is volume is given as:

ΔV(VgraphiteVdiamond)\Delta V({V_{graphite}} - {V_{diamond}})

Now we will find the volume of diamond and volume of graphite, As we know Density=MassVolumeDensity = \dfrac{{Mass}}{{Volume}}

Given that the density of diamond=3g/cc3g/cc and the density of graphite is 2g/cc2g/cc. On putting the value of density in the above formula we get:

3=12volume of diamond3 = \dfrac{{{\text{12}}}}{{{\text{volume of diamond}}}}

volume of diamond=123=4cm3 \Rightarrow {\text{volume of diamond}} = \dfrac{{12}}{3} = 4c{m^3}

Similarly

2=12volume of graphite2 = \dfrac{{{\text{12}}}}{{{\text{volume of graphite}}}}

volume of graphite=122=6cm3 \Rightarrow {\text{volume of graphite}} = \dfrac{{12}}{2} = 6c{m^3}

Now to calculate the change is volume, we will find the change in (ΔV\Delta V) using the below formula:ΔV=VgraphiteVdiamond\Delta V = {V_{graphite}} - {V_{diamond}}

ΔV=64=2cm3 \Rightarrow \Delta V = 6 - 4 = 2c{m^3}

On putting the value of ΔV\Delta VIn equation 1, we get:

ΔH=ΔU+5×105bar×2×103lt/mol\Delta H = \Delta U + 5 \times {10^5}bar \times 2 \times {10^{ - 3}}lt/mol

ΔH=ΔU+103 \Rightarrow \Delta H = \Delta U + {10^3}

ΔHΔU=103 \Rightarrow \Delta H - \Delta U = {10^3}

Taking negative sign common, the above equation becomes:

ΔUΔH=103bar Ltr \Rightarrow \Delta U - \Delta H = - {10^3}bar{\text{ L}}tr

As we know 1 bar ltr=100J1{\text{ }}bar{\text{ }}ltr = 100J

Then 103bar ltr=1000×100J/mol - {10^3}bar{\text{ }}ltr = - 1000 \times 100J/mol

Or

103bar ltr=100kJ/mol {10^3}bar{\text{ }}ltr = 100kJ/mol

Hence the correct answer is option (A).

Note:

Both graphite and diamonds are made of carbon and are known as allotropes of carbon, i.e. the arrangement of carbon atoms in space is different. The atoms of carbon in a diamond are arranged tetrahedrally and for the case of graphite, they are arranged in an infinite array.