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JEE · NEET Previous Year Question

$An ideal gas occupies a volume of 2 m^{3} at a pressure of 3 \times 10^{6} Pa. The energy of the gas :$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $9 \times 10^{6} J$
B $6 \times 10^{4} J$
C $10^{8} J$
D $3 \times 10^{2} J$

Correct answer: A

Solution

(a) JEE Main 2019 Energy of the gas, E = \frac{f}{2} nRT = \frac{f}{2} PV = \frac{f}{2} (3 \times 10^{6}) (2) = f \times 3 \times 10^{6} Considering gas is monoatomic i.e., f = 3 Energy, E = 9 \times 10^{6} J

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