Engineering Thermodynamics In a no flow reversible process for which p = (-3V + 15) × 105 N/m², V changes from 1 m3 to 2 m3. The work done will be about 10 × 10⁵ kilo joules 100 × 10⁵ joules 1 × 10⁵ joules 10 × 10⁵ joules 10 × 10⁵ kilo joules 100 × 10⁵ joules 1 × 10⁵ joules 10 × 10⁵ joules ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics For reversible adiabatic process, change in entropy is Zero Minimum Negative Maximum Zero Minimum Negative Maximum ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics An adiabatic wall is one which Permits thermal interaction Encourages thermal interaction Prevents thermal interaction Discourages thermal interaction Permits thermal interaction Encourages thermal interaction Prevents thermal interaction Discourages thermal interaction ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics On volume basis, air contains following parts of oxygen 77 21 23 25 77 21 23 25 ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics Work done is zero for the following process All of the listed here Free expansion Throttling Constant volume All of the listed here Free expansion Throttling Constant volume ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics The hyperbolic process is governed by Charles' law Boyle's law Avogadro's law Gay-Lussac law Charles' law Boyle's law Avogadro's law Gay-Lussac law ANSWER DOWNLOAD EXAMIANS APP
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