Question 792: Thermodynamics & KTG

Question

A diatomic gas undergoes adiabatic change. Its pressure P and temperature T are related as $\mathrm{P} \propto \mathrm{T}^{\mathrm{x}}$ where the value of x is

Accepted Answer

For a diatomic gas undergoing an adiabatic process, the relationship between pressure ($P$) and temperature ($T$) can be derived using the adiabatic condition and the gas law.
The adiabatic condition for an ideal gas is given by:
$$ PV^{\gamma} = 	ext{constant} $$
where $P$ is pressure, $V$ is volume, and $\gamma$ (gamma) is the adiabatic index or heat capacity ratio given by:
$$ \gamma = \frac{C_p}{C_v} $$
For a diatomic gas, the degrees of freedom $ f $ are 5, so:
$$ C_v = \frac{f}{2}R = \frac{5}{2}R $$
$$ C_p = C_v + R = \frac{7}{2}R $$
Thus,
$$ \gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} $$
The ideal gas law is given by:
$$ PV = nRT $$
During an adiabatic process, we also have:
$$ P \propto V^{-\gamma} $$
and
$$ T \propto V^{\gamma-1} $$
Substituting the relationship for temperature:
From the ideal gas law $PV = nRT$, we have $P = \frac{nRT}{V}$.
We can substitute $T \propto V^{\gamma-1}$ into the ideal gas law:
$$ P \propto \left(V^{\gamma-1}ight)^{\frac{1}{V}} \Rightarrow P \propto \left(V^{1-\gamma}ight) $$
Hence, $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Plugging in the value of ÃÂ³ for a diatomic gas:
$$ \gamma = \frac{7}{5} $$
$$ \frac{\gamma}{\gamma-1} = \frac{\frac{7}{5}}{\frac{7}{5}-1} = \frac{\frac{7}{5}}{\frac{2}{5}} = \frac{7}{2} $$
Thus, for a diatomic gas, the relationship is:
$$ P \propto T^{\frac{7}{2}} $$
Therefore, the value of $ x $ is $ 3.5 $.
Answer:
Option A, $ x = 3.5 $.

Answer

3.5

Answer

2.5

Answer

4.5

Answer

3

Thermodynamics & KTG

A diatomic gas undergoes adiabatic change. Its pressure P and temperature T are related as \mathrm{P} \propto \mathrm{T}^{\mathrm{x}} where the value of x is

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