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JEE PHYSICS MCQ: THERMODYNAMICS

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The change in internal energy of a given mass of gas, when its volume changes fromVV to 2VV at constant pressure pp is (CpCv=Ξ³(\frac{{{C_p}}}{{{C_v}}} = \gamma universal gas constant=R)
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Correct option: D
pV(Ξ³βˆ’1)\frac{{pV}}{{\left( {\gamma - 1} \right)}}
Option D (pV(Ξ³βˆ’1)\frac{{pV}}{{\left( {\gamma - 1} \right)}}) is correct because the change in internal energy is calculated using the formula Ξ”U=nCVΞ”T\Delta U = nC_V\Delta T 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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