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# If work done by an ideal diatomic gas in a process from $\mathrm{A}$ to $\mathrm{B}$ as shown in $\mathrm{P}-\mathrm{V}$ graph is $20 \mathrm{J}$ :Then increase in internal energy of the gas is:(A) $40 \mathrm{J}$(B) $60 \mathrm{J}$(C) $100 \mathrm{J}$(D) $120 \mathrm{J}$

Last updated date: 20th Jun 2024
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Hint: We know that the internal energy is an extensive property: it depends on the size of the system, or on the amount of substance it contains. At any temperature greater than absolute zero, microscopic potential energy and kinetic energy are constantly converted into one another, but the sum remains constant in an isolated system. The internal energy of a system can be understood by examining the simplest possible system: an ideal gas. Because the particles in an ideal gas do not interact, this system has no potential energy. The internal energy of an ideal gas is therefore the sum of the kinetic energies of the particles in the gas.

$\mathrm{W}=$ Area $=\dfrac{1}{2} \mathrm{P} \mathrm{V}=\dfrac{1}{2} \mathrm{nR} \Delta \mathrm{T}=20 \mathrm{J}$
Now the change in internal energy is given as $\dfrac{\mathrm{f}}{2} \mathrm{nR} \Delta \mathrm{T}$ (here $\mathrm{f}$ is the number of degree of freedom)
For a diatomic gas $f=5 .$ Thus we get $\Delta \mathrm{U}=\dfrac{5}{2} \mathrm{nR} \Delta \mathrm{T}=100 \mathrm{J} .=\mathrm{x}$