
For an adiabatic compression the quality $P{V^\gamma }$.
(A) Increases
(B) Decreases
(C) Remains constant
(D) Depends on $\gamma$
Answer
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Hint The compression in which there is no heat added or subtracted from the air and the internal energy of the air is increased by an amount which is equivalent to the external work on the air. Based on this concept, we can solve this question.
Complete step-by-step answer:
We know that equation of state is given by:
$T{V^{Y - 1}} =$constant
${T_1}{V_1}^{Y - 1} = {T_2}{V_2}^{Y - 1}$
$\dfrac{{{T_2}}}{{{T_1}}} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{Y - 1}}$
When the gas gets compressed, the temperature thus increases.
$PV = nRT\;$
${P_2}{V_2} - {P_1}{V_1} = nR({T_2} - {T_1})$
Since temperature increases, PV also increases.
Hence, the correct answer is Option A.
Note It should be known that the adiabatic compression of the gas causes a rise in the temperature of the gas. In the case of adiabatic expansion against pressure, or a spring, will result in the drop of the temperature. In this contrast, the free expansion is defined as an isothermal process for an ideal gas.
Complete step-by-step answer:
We know that equation of state is given by:
$T{V^{Y - 1}} =$constant
${T_1}{V_1}^{Y - 1} = {T_2}{V_2}^{Y - 1}$
$\dfrac{{{T_2}}}{{{T_1}}} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{Y - 1}}$
When the gas gets compressed, the temperature thus increases.
$PV = nRT\;$
${P_2}{V_2} - {P_1}{V_1} = nR({T_2} - {T_1})$
Since temperature increases, PV also increases.
Hence, the correct answer is Option A.
Note It should be known that the adiabatic compression of the gas causes a rise in the temperature of the gas. In the case of adiabatic expansion against pressure, or a spring, will result in the drop of the temperature. In this contrast, the free expansion is defined as an isothermal process for an ideal gas.
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