Question

Using the first law of the thermodynamics for an adiabatic processes, establish the relation PV power of gama =K Where gama is the ratio pf the heat capacity of constant pressure to that at constant volume .plot this equation on a P-v diagram .What will be its slope?

Answer 1

Answer on Question #66810, Physics / Molecular Physics | Thermodynamics

Using the first law of the thermodynamics for an adiabatic processes, establish the relation PV power of gama =K. Where gama is the ratio pf the heat capacity of constant pressure to that at constant volume. plot this equation on a P-v diagram. What will be its slope?

Solution:

According to the first law of thermodynamics,

$$\mathrm{d}U + \delta W = \delta Q = 0$$

where $dU$ is the change in the internal energy of the system and $\delta W$ is work done by the system. Any work $(\delta W)$ done must be done at the expense of internal energy $U$, since no heat $\delta Q$ is being supplied from the surroundings.

Pressure-volume work $\delta W$ done by the system is defined as

$$\delta W = P dV$$

However, $P$ does not remain constant during an adiabatic process but instead changes along with $V$. It is desired to know how the values of $dP$ and $dV$ relate to each other as the adiabatic process proceeds. For an ideal gas the internal energy is given by

$$U = \alpha nRT$$

where $\alpha$ is the number of degrees of freedom divided by two, $R$ is the universal gas constant and $n$ is the number of moles in the system (a constant).

Differentiating Equation (3) and use of the ideal gas law, $PV = nRT$, yields

$$U = \alpha nRT = \alpha d(PV) = \alpha (PdV + VdP)$$

Equation (4) is often expressed as $dU = nC_V dT$ because $C_V = \alpha R$.

Now substitute equations (2) and (4) into equation (1) to obtain

$$- PdV = \alpha PdV + \alpha V dP (PdV + VdP)$$

factorize $-P\,dV$:

$$- (\alpha + 1) PdV = \alpha V dP$$

and divide both sides by $PV$:

$$- (\alpha + 1) \frac{dV}{V} = \alpha \frac{dP}{P}$$

After integrating the left and right sides from $V_0$ to $V$ and from $P_0$ to $P$ and changing the sides respectively,

$$\ln \frac{P}{P_0} = \frac{\alpha + 1}{\alpha} \ln \frac{V}{V_0}$$ $$\text{Of} \Rightarrow \frac{P}{P_0} = \left(\frac{V}{V_0}\right)^{-\gamma}$$

where $\gamma = \frac{\alpha + 1}{\alpha}$

Therefore,

$$\left(\frac {P}{P _ {0}}\right) \times \left(\frac {V}{V _ {0}}\right) ^ {v} = 1 (1 0)$$

Of (10) $\Rightarrow$ P0V0 = PVv = const

Adiabat on PV diagram is a "steep hyperbola",

the slope is a tangent of angle,

tangent of angle for adiabat is greater than tangent of angle for isotherms (adiabat is steeper isotherms)

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