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JEE PHYSICS MCQ: THERMODYNAMICS

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An ideal gas undergoing adiabatic change has the following pressure-temperature relationship
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Correct option: D
P1βˆ’Ξ³TΞ³=P^{1-\gamma} T^\gamma= constant
The correct answer is P1βˆ’Ξ³TΞ³=P^{1-\gamma} T^{\gamma}= constant because this relationship is derived from the adiabatic process equation and the ideal gas law.

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The work done in which of the following process is zero? Water is used to cool the radiators of engines in cars because A gas expand 0.25  m30.25\,\,{m^3} at constant pressure 103N/m2,{10^3}N/{m^2}, the work done is A thermodynamic process in which temperature TT of the system remains constant though other variable PP and VV may change, is called What is the slope for an isothermal process? For adiabatic processes (Ξ³=CpCv)\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)
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