An Ideal Gas Occupies A Volume Of 2M3 At A Pressure Of 3×10 6Pa. The Energy Of The Gas Is:

For an ideal gas, the internal energy is given as:

\(E = \frac{f}{2}nRT\) \(\Rightarrow \frac{3}{2}nRT\)

As [pV=nRT] \(E = \frac{3}{2}pV\)

Where:

p = pressure of the gas= \(3 * 10^{6}Pa\)

V = volume of the gas= \(2 m^{3}\)

By substituting the given values, we get

\( E = \frac{3}{2}pV\) \( \Rightarrow \frac{3}{2} * 3 * 10^{6}Pa * 2 m^{3}\) \( \Rightarrow 9 * 10^{6}J\)

Therefore, the energy of the gas is \( 9 * 10^{6}J\)

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