InterviewSolution
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InterviewSolution
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At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this, at higher temperatureA. `C_(V) = (3 R)/(2)` for a monatomic gasB. `C_(V) gt (3 R)/(2)` for a monatomic gasC. `C_(V) lt (3 R)/(2)` for a diatomic gasD. `C_(V) gt (3 R)/(2)` for a diatomic gas |
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Answer» Correct Answer - A::D Vibrational kinetic energy of a monatomic gas `=0` at all temperature. So, `C_(v)=3 R//2` for a monoatomic gas at high temperatures also. In case of a diatomic gas `C_(v)=5 R//2` at low temperatures while , `C_(v)=5R//2` at high temperatures due to vibrational `KE`. |
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