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2. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio C_{p}/C_{V} for the gas is

(A) 4/3

(B) 2

(C) 5/3

(D) 3/2

Ans A.

4.

One mole of ideal monoatomic gas ( C_{p}/C_{V}= 5/30) is mixed with one mole of diatomic gas (C_{p}/C_{V} = 7/5). What is C_{p}/C_{V} for the mixture?

(A) 3/2

(B) 23/15

(C) 35/23

(D) 4/3

Ans A.

5 If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be

(A) 4

(B) 16

(C) 32

(D) 64.

I WANT TO KNOW HOW TO SOLVE IT!!

9 years ago

Daer Aadithya,

Basically the symbol δ means "a little change in," like δA means "a little change in A." Try to keep up with this, ok?

The most important equation in thermodynamics can be summarized in one neat little form, the first law of thermodynamics. It says that reversible processes obey the equation:

δE = T δS − P δV + Σ? μ? δN?

It says that the internal energy of a set of particles in a box changes only if the volume of the box changes, the entropy of the particles changes, or the number of particles of a given type N? changes. These changes relate to the internal energy by certain thermodynamic quantities: For entropy, it is temperature; for volume, it is pressure, for particle number, it is the chemical potential of that particle. Systems which are allowed to trade volume, particles, or energy will equalize their pressure, chemical potential, or temperature, respectively.

This tells you that the internal energy E does not change if the entropy S, the volume V, and the particle numbers N? are not changing. Thus E is a good "thermodynamic potential" for describing the energy available at constant volume.

The first law in its form is also helpful to understand the "Legendre transforms" of E. The easiest example is when we want to look at a system with constant pressure, rather than constant volume. This works like so: you have two systems with volumes V1 and V2, and in a constant pressure situation, V2 is so much larger than V1 that any change in the volume of V1 does not change the pressure of V2 much at all. You may, if you like, imagine that V2 is the atmosphere of the Earth while V1 is something which is rapidly expanding -- a balloon or so -- within that atmosphere. Since the atmosphere sets a common pressure, we would like to use a different "thermodynamic potential" which tells us how much energy we can get, given that this thing will change in volume to keep a constant pressure.

This is not so hard as it looks. You just introduce the Enthalpy:

H = E + P V

...and then take its differential with the product rule from calculus:

δH = δE + P δV + V δP

(Notice that δ(P V) = (P + δP) (V + δV) - P V = P δV + V δP + δP δV. The last term is very very tiny if both δP and δV are tiny, so we ignore it completely. Thats very common in calculus.)

...we can then use the thermodynamic identity for δE, canceling the terms ± P δV:

δH = T δS + V δP + Σ? μ? δN?

Thus the enthalpy remains unchanged as long as entropy, pressure, and particle

numbers are unchanged. At constant pressure, the enthalpy is the right

"thermodynamic potential" for calculating energy differences. For an ideal gas:

P V = n R T

E = α n R T

The specific heat at constant volume is just:

C_v = (1/n) (∂E/∂T) = α R

But the specific heat at constant pressure is more like:

H = E + P V = (α + 1) n R T

C_p = (1/n) (∂H/∂T) = (α + 1) R.

Thus there is a useful number to know called the specific heat ratio:

γ = C_p / C_v = (α + 1)/α = 1 + 1/α.

Why is this useful? Well, it enters into the formula for an adiabatic process:

adiabatic process: δS = 0, δN? = 0

δE + P δV = 0

This quantity δE + P δV is sometimes called δQ, the "heat flow", which is why

this is called an adiabatic ("constant heat") process. It is a fundamentally

reversible process due to δS = 0. Using the ideal gas equation gives:

E = α P V,

δE = α (P δV + V δP)

(α + 1) P δV + α V δP = 0

(α + 1) δV/V = - α δP/P

Integrating gives:

γ log(V) = -log(P) + log(C)

V^γ = C / P

P V^γ = constant.

This is the equation for any adiabatic process.

For ideal gases, PV = n R T, and thus for adiabatic processes on ideal gases:

−

P V^γ * (n R T / P V)^γ = constant

If we slide the (n R)^γ into the constant we can write this as:

T / P^(γ − 1) = constant.

Using that γ − 1 = 1/α, we raise both of these to the exponent α and use exponent rules to find:

T^α = constant * P.

Thus when they tell us P ~ T³, that means α = 3, so that E = 3 n R T. In turn:

C_p / C_v = (α + 1)/α = 4/3.

Best Of luck

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