Thermodynamics

JEE Main level, numerical answer, medium

116 of 190

Question 116 of 190

The total internal energy of two mole monoatomic ideal gas at temperature T=300KT=300K will be _____ JJ. (Given R=8.31Jmol−1⋅KR=8.31J{mol}^{-1}\cdot K)

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Answer: 7479

The internal energy of an ideal gas is a state function and is given by, U=nCvTU=n{C}_{v}T.
For monatomic gas, Cv=32R{C}_{v}=\dfrac{3}{2}R.
Therefore, U=2×32×8.31×300=7479JU=2\times \dfrac{3}{2}\times 8.31\times 300=7479J

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