# 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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