An ideal gas occupies a volume of 2 m 3 at a pressure of 3 × 10 6 P a . The energy of the gas is:

V = 2 m 3 P = 3 × 10 6 P a P V = n R T E = f 2 n R T = f 2 P V = 3 2 × 3 × 10 6 × 2 = 9 × 10 6 J

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PhysicsHeat & ThermoKTGEasy2 minPYQ_2019

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

Options:

Answer:

Solution:

$V = 2 m^{3}$

$P = 3 \times 10^{6} P a$

$P V = n R T$

$E = \frac{f}{2} n R T = \frac{f}{2} P V = \frac{3}{2} \times 3 \times 10^{6} \times 2$

$= 9 \times 10^{6} J$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

⏱ 2mℹ️ Source: PYQ_2019

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