In isobaric process of ideal gas (f=5) work done by gas is equal of 10J. Then heat given to gas during process:
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Answer:
An Isobaric process is a thermodynamic process in which the pressure stays constant: ΔP = 0. The heat transferred to the system does work, but also changes the internal energy of the system. This article uses the physics sign convention for work, where positive work is work done by the system. Using this convention, by the first law of thermodynamics 
where W is work, U is internal energy, and Q is heat.[1] Pressure-volume work by the closed system is defined as:
{\displaystyle W=\int \!p\,dV\,}
where Δ means change over the whole process, whereas d denotes a differential. Since pressure is constant, this means that
{\displaystyle W=p\Delta V\,}.
Applying the ideal gas law, this becomes
{\displaystyle W=n\,R\,\Delta T}
assuming that the quantity of gas stays constant, e.g., there is no phase transition during a chemical reaction. According to the equipartition theorem,[2] the change in internal energy is related to the temperature of the system by
{\displaystyle \Delta U=n\,C_{P}\,\Delta T},
where cP is specific heat at a constant pressure.
Substituting the last two equations into the first equation produces:
{\displaystyle {\begin{aligned}Q&=n\,C_{P}\,\Delta T-n\,R\,\Delta T\\Q&=n\Delta T(C_{P}-R)\end{aligned}}}