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The pressure and density of a diatomic gas $\left(\gamma=\dfrac{7}{5}\right)$ change adiabatically from $\left(P_{1}, \rho_{1}\right)$ to $\left(P_{2}, \rho_{2}\right) .$ If $\dfrac{\rho_{2}}{\rho_{1}}=32,$ then $\dfrac{P_{2}}{P_{1}}$ should be
A. 126
B. 128
C. 146
D. 124

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Last updated date: 13th Jun 2024
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Answer
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Hint: An adiabatic cycle happens without moving heat or mass between a thermodynamic framework and its environmental factors. In contrast to an isothermal cycle, an adiabatic cycle moves vitality to the environmental factors just as work. It additionally thoughtfully undergirds the hypothesis used to elucidate the principal law of thermodynamics and is in this manner a key thermodynamic concept.

Complete step by step solution:
We know that for an adiabatic process,
$\mathrm{PV}^{\gamma}=\mathrm{constant}$
$\mathrm{P} \rho^{-\gamma}=$ constant
$\dfrac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\dfrac{\rho_{1}}{\rho_{2}}\right)^{-\gamma}=\left(\dfrac{1}{32}\right)^{-7 / 5}=(32)^{7 / 5}=128$

Therefore, the correct answer is Option B.

Note: Thermodynamics is the part of material science that manages the connections among heat and different types of vitality. Specifically, it depicts how warm vitality is changed over to and from different types of vitality and how it influences matter.