State the first law of thermodynamics. Establish the relation between Cp and Cv
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The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
Thermodynamics is a branch of physical chemistry that describes the relationship of heat energy with other forms of energy. CV and CP are two terms used in thermodynamics. CV is the specific heat at constant volume, and CP is the specific heat at constant pressure. Specific heat is the heat energy required to raise the temperature of a substance (per unit mass) by one degree Celsius. The main difference between CV and CP is that the volume change is zero for a system under CV whereas the pressure change is zero for a system under CP.
Thermodynamics is a branch of physical chemistry that describes the relationship of heat energy with other forms of energy. CV and CP are two terms used in thermodynamics. CV is the specific heat at constant volume, and CP is the specific heat at constant pressure. Specific heat is the heat energy required to raise the temperature of a substance (per unit mass) by one degree Celsius. The main difference between CV and CP is that the volume change is zero for a system under CV whereas the pressure change is zero for a system under CP.
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Hi,
Here is your answer,
FIRST LAW OF THERMODYNAMICS
The quantity of heat energy of heat energy absorbed by the system is equal to the sum of increases in the internal energy of the system and the external work done by it.
dQ = du + dw
But, dw = Pdv
∴ dQ = du + Pdv
{ SIGN CONVENTION }
(1) Heat is absorbed by the system dQ = +ve
(2) Heat is released by the system dQ = +ve
(3) Work done by the system dw = +ve
(4) Work done on the system dw = -ve
(5) Increase on internal Energy du = +ve
(6) Decrease in Internal Energy du = -ve
Now, Let us consider a gas with a massless and frictionless pistion. Let the gas expands by a very small volume dV.
Let p is pressure and V is volume of the gas. A is the area of cross-section of the piston, where dx is the distance moved by the piston. Let the volume expansion is dV.
So, the work done by the gas
dW = force × dz
dW = pA × dx = p( A dx) = pdV [ ∴ Adx = dV]
Now, dW = pdV
For small displacement dx, the pressure is assumed to be constant.
Thus,from first law ΔQ = ΔU + pdV
where,
ΔQ = heat transition between system and surrounding
ΔU = change in internal energy of the system.
{ RELATION BETWEEN Cp and Cv }
→ We called this relation as Mayer's Formula.
We can establish the relation between specific heat capacity at constant volume (Cv) and specific heat capacity at constant pressure (Cp) of a gas.
For an ideal gas, the relation between Cp and Cv is
Cp - Cv = R -------------------> (1)
∴ This relation is known as Mayer's Formula.
Now, To establish the relation, we need to begin from the first law of thermodynamics for 1 mole of gas.
ΔQ = ΔU + pΔV
If heat ΔQ is absorbed at constant volume,
∴ pΔV = 0 and ΔQ = CvΔT for one mole of a gas
Now, ΔV = 0
Then, Cv = (ΔQ/ΔT)v = (ΔU/ΔT)v = (ΔU/ΔT) -----------------> (2)
where the V is dropped in the last step, since U of an ideal gas depends only on the temperature, not on the volume.
Now, heat ΔQ is absorbed at constant pressure, then
ΔQ = CpΔT
Cp = (ΔQ/ΔT)p = (ΔU/ΔT)p = (ΔU/ΔT)p
Now, p can be the dropped from the first term since U of an ideal gas depends only on T, not on pressure.
Now, by using Eq. (2)
or Cp = Cv +p(ΔV/Δp)p --------------------> (3)
Now, for 1 mole of an ideal gas, PV = RT
If the pressure is kept constant
p(ΔV/ΔT)p = R -----------------------> (4)
From Eq. (2) , (3) and (4)
{ Cp - Cv = R }
Here, Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure and volume and R is the universal gas constant.
The ratio of Cp and Cv is notified as γ
γ = Cp/Cv
And it is also known as heat capacity ratio,
Cv = R/r-1 and Cp = γ R/r-1
Hope it helps you !
Here is your answer,
FIRST LAW OF THERMODYNAMICS
The quantity of heat energy of heat energy absorbed by the system is equal to the sum of increases in the internal energy of the system and the external work done by it.
dQ = du + dw
But, dw = Pdv
∴ dQ = du + Pdv
{ SIGN CONVENTION }
(1) Heat is absorbed by the system dQ = +ve
(2) Heat is released by the system dQ = +ve
(3) Work done by the system dw = +ve
(4) Work done on the system dw = -ve
(5) Increase on internal Energy du = +ve
(6) Decrease in Internal Energy du = -ve
Now, Let us consider a gas with a massless and frictionless pistion. Let the gas expands by a very small volume dV.
Let p is pressure and V is volume of the gas. A is the area of cross-section of the piston, where dx is the distance moved by the piston. Let the volume expansion is dV.
So, the work done by the gas
dW = force × dz
dW = pA × dx = p( A dx) = pdV [ ∴ Adx = dV]
Now, dW = pdV
For small displacement dx, the pressure is assumed to be constant.
Thus,from first law ΔQ = ΔU + pdV
where,
ΔQ = heat transition between system and surrounding
ΔU = change in internal energy of the system.
{ RELATION BETWEEN Cp and Cv }
→ We called this relation as Mayer's Formula.
We can establish the relation between specific heat capacity at constant volume (Cv) and specific heat capacity at constant pressure (Cp) of a gas.
For an ideal gas, the relation between Cp and Cv is
Cp - Cv = R -------------------> (1)
∴ This relation is known as Mayer's Formula.
Now, To establish the relation, we need to begin from the first law of thermodynamics for 1 mole of gas.
ΔQ = ΔU + pΔV
If heat ΔQ is absorbed at constant volume,
∴ pΔV = 0 and ΔQ = CvΔT for one mole of a gas
Now, ΔV = 0
Then, Cv = (ΔQ/ΔT)v = (ΔU/ΔT)v = (ΔU/ΔT) -----------------> (2)
where the V is dropped in the last step, since U of an ideal gas depends only on the temperature, not on the volume.
Now, heat ΔQ is absorbed at constant pressure, then
ΔQ = CpΔT
Cp = (ΔQ/ΔT)p = (ΔU/ΔT)p = (ΔU/ΔT)p
Now, p can be the dropped from the first term since U of an ideal gas depends only on T, not on pressure.
Now, by using Eq. (2)
or Cp = Cv +p(ΔV/Δp)p --------------------> (3)
Now, for 1 mole of an ideal gas, PV = RT
If the pressure is kept constant
p(ΔV/ΔT)p = R -----------------------> (4)
From Eq. (2) , (3) and (4)
{ Cp - Cv = R }
Here, Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure and volume and R is the universal gas constant.
The ratio of Cp and Cv is notified as γ
γ = Cp/Cv
And it is also known as heat capacity ratio,
Cv = R/r-1 and Cp = γ R/r-1
Hope it helps you !
Anonymous:
mark as brainliest answer !
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