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Question: In an insulated container \(1\) mole of a liquid. Molar volume \(100\) ml at 1 bar. Liquid is steepl...

In an insulated container 11 mole of a liquid. Molar volume 100100 ml at 1 bar. Liquid is steeply taken to 100100 bar, when volume of the liquid decreases by 11ml. find ΔH\Delta {\rm H} for the process.
A.79007900 bar ml
B.89008900 bar ml
C.99009900 bar ml
D.1090010900 bar ml

Explanation

Solution

Enthalpy of the chemical reaction is defined as the sum of the internal energy and the product of its pressure and volume which is expressed as H=U+pΔVH = U + p\Delta V. Change in the volume of the chemical reaction at constant pressure is referred to as pressure-volume work.

Complete answer:
From the first law of thermodynamics we know the relation between internal energy, heat and work of the system.
ΔU=q+w\Delta U = q + w
Since, an insulated chamber is used in the reaction which means the reaction is conducted under adiabatic conditions where heat change becomes zero (q=0)\left( {q = 0} \right).
ΔU=w\Delta U = w……….(i)\left( i \right)
We know that work depend on the pressure and volume of reaction which is expressed as-
w=−p(V2−V1)w = - p\left( {{V_2} - {V_1}} \right)
Where, is pressure of system which is 100100 in reaction
Is final volume of system which is 9999 ml
Is initial volume of system which is 100100 ml
w=−100(99−100)w = - 100\left( {99 - 100} \right)
w=−100(−1)w = - 100\left( { - 1} \right)
After calculating this reaction, we get
w=100w = 100
After putting the value of work in the equation (i)\left( i \right)
ΔU=w\Delta U = w
ΔU=100\Delta U = 100 bar ml
From the equation of enthalpy as given above –
H=U+ΔpVH = U + \Delta pV
ΔH=ΔU+(p2V2−p1V1)\Delta H = \Delta U + \left( {{p_2}{V_2} - {p_1}{V_1}} \right)
Put all the obtained values in the above equation to obtain the value of ΔH\Delta {\rm H}.
ΔH=100+(100×99−1×100)\Delta {\rm H} = 100 + \left( {100 \times 99 - 1 \times 100} \right)
ΔH=100+(9900−100)\Delta {\rm H} = 100 + \left( {9900 - 100} \right)
After solving the above equation, we get,
ΔH=100+9800\Delta {\rm H} = 100 + 9800
Hence, ΔH=9900\Delta {\rm H} = 9900 bar ml. Therefore option C is the correct option.

Note:
Amount of heat absorbed at constant volume is expressed as ΔU\Delta U as while the amount of heat absorbed at constant pressure is expressed as ΔH\Delta {\rm H}.
Take the unit of pressure in the bar and volume of solution in ml.