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JEE Main 202228 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
28 June
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2022 · 28 June · Morning Shift · Physics · Thermodynamics

The total internal energy of two mole monoatomic ideal gas at...

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)

Correct answer

0

Step-by-step solution

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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