Thermodynamics

JEE Main level, numerical answer, medium

110 of 190

Question 110 of 190

At a certain temperature, the degrees of freedom per molecule for gas is 88. The gas performs 150J150J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be _____ JJ.

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Answer: 750

For constant pressure process,
Work done is given by W=nRΔT=150JW=nR\Delta T=150J, here, RR is gas constant.
Now, for above process, f=8f=8.
So, specific heat at constant pressure is CP=(1+f2)R{C}_{P}=\left(1+\dfrac{f}{2}\right)R.
Amount of heat adsorbed by the gas is
Q=nCPRΔT=(f2+1)nRΔT=(82+1)150=750JQ=n{C}_{P}R\Delta T=\left(\dfrac{f}{2}+1\right)nR\Delta T=\left(\dfrac{8}{2}+1\right)150=750J

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