The work done equals \(\smallint pdv\) for
Non–flow quasi-static process
In a non-flow quasi-static process, the pressure at the moving boundary is well defined, so boundary work is \(W_b=\int_{V_1}^{V_2}p\,dV\). For specific volume v, the corresponding specific work is \(w_b=\int p\,dv\). Quasi-static motion alone does not guarantee full reversibility; friction can still be present. Paddlewheel work is not boundary work.
Important Questions from First Law of Thermodynamics
Heat transfer in a cyclic process are +20 kJ, -5 kJ, -10 kJ and +15kJ. Net work done for this cycle will be given by:
An open system is one in which ______.
A system will be in thermodynamic equilibrium only if it is in ____.
A. Thermal equilibrium
B. Mechanical equilibrium
C. Chemical equilibrium
The gas constant (R) is equal to the _____ of two specific heats.
In a reversible adiabatic process the ration \(\left( {\frac{{{T_1}}}{{{T_2}}}} \right)\) is equal to:
