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Question:

An ideal gas occupies a volume of 2m³ at a pressure of 3×10⁶Pa. The energy of the gas is:

3×10²J

108J

9×10⁶J

6×10⁴J

Solution:

Energy = 1/2nRT = f/2PV = f/2(3×10⁶)(2) = f × 3×10⁶
Considering gas is monoatomic i.e. f=3
E = 9×10⁶J