Question 755: Thermodynamics & KTG

Question

In an adiabatic process for an ideal gas, the relation between the universal gas constant ' $R$ ' and specific heat at constant volume ' $\mathrm{C}_{\mathrm{v}}$ ' is $R=0.4 C_v$. The pressure ' $P$ ' of the gas is proportional to the temperature ' $T$ ', of the gas as $T^k$. The value of constant ' K ' is

Accepted Answer

In an adiabatic process for an ideal gas, where the universal gas constant $ R $ is related to the specific heat at constant volume $ \mathrm{C}_\mathrm{v} $ as $ R = 0.4 \mathrm{C}_\mathrm{v} $, the pressure $ P $ is proportional to the temperature $ T $ raised to the power $ K $, or $ P \propto T^K $.
For such a process, the relationship can be expressed as:
$ \mathrm{PT}^{-K} = 	ext{constant} $
In an adiabatic process for an ideal gas, another important relationship is:
$ \mathrm{PT}^{\frac{\gamma}{1-\gamma}} = 	ext{constant} $
From these, we deduce:
$ \frac{\gamma}{1-\gamma} = -K \quad 	ext{.... (i)} $
Furthermore, the specific heat at constant pressure $ \mathrm{C}_\mathrm{p} $ can be expressed as:
$ \mathrm{C}_\mathrm{p} = \mathrm{C}_\mathrm{v} + R $
$ \mathrm{C}_\mathrm{p} = \mathrm{C}_\mathrm{v} + 0.4 \mathrm{C}_\mathrm{v} $
$ \mathrm{C}_\mathrm{p} = 1.4 \mathrm{C}_\mathrm{v} $
The adiabatic index $ \gamma $ is given by:
$ \gamma = \frac{\mathrm{C}_\mathrm{p}}{\mathrm{C}_\mathrm{v}} = 1.4 \quad	ext{... (ii)} $
By combining (i) and (ii), we have:
$ \frac{\gamma}{1-\gamma} = \frac{1.4}{1-1.4} = \frac{-1.4}{-0.4} = \frac{-14}{4} = \frac{-7}{2} $
Thus, solving for $ K $:
$ \frac{-7}{2} = -K $
$ K = \frac{7}{2} $

Answer

$\frac{7}{2}$

Answer

$\frac{7}{3}$

Answer

5

Answer

5

Thermodynamics & KTG

In an adiabatic process for an ideal gas, the relation between the universal gas constant ' R ' and specific heat at constant volume ' \mathrm{C}_{\mathrm{v}} ' is R=0.4 C_v. The pressure ' P ' of the gas is proportional to the temperature ' T ', of the gas as T^k. The value of constant ' K ' is

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