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An ideal gas occupies a volume of $2{m^3}$ at a pressure of $3 \times {10^6}pa$ the energy of the gas is:
A) $3 \times {10^2}J$
B) ${10^3}J$
C) $6 \times {10^4}J$
D) $9 \times {10^6}J$

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Last updated date: 20th Apr 2024
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Answer
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Hint: Internal energy is the sum of the kinetic and potential energies of a particle. As the particles in an ideal gas do not interact, the system has no potential energy. So the internal energy of an ideal gas is, therefore, the sum of the kinetic energies of the particles in the gas.

Formulae used:
$E = \dfrac{3}{2}PV$
Where $E$ is the energy of the ideal gas, $P$ is the pressure of the ideal gas and $V$ is the volume of the ideal gas.

Complete step by step answer:
In the question, the volume of the ideal gas is given as
$ \Rightarrow V = 2{m^3}$
And the pressure of the ideal gas is
$ \Rightarrow P = 3 \times {10^6}pa$
As $1pa = 1\dfrac{N}{{{m^2}}}$
So the pressure becomes
$ \Rightarrow P = 3 \times {10^6}\dfrac{N}{{{m^2}}}$
For any idea gas, the energy is given by,
$ \Rightarrow E = \dfrac{3}{2}PV$
Where $E$ is the energy of the ideal gas, $P$ is the pressure of the ideal gas and $V$ is the volume of the ideal gas.
Substituting the values of $P$ and $V$ we get,
$ \Rightarrow E = 9 \times {10^6}J$

So the answer is an option (D).

Additional Information: An ideal gas is a theoretical gas made up of many atoms that is not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law. At standard temperature and pressure conditions, most real gases behave like an ideal gas. Many gases such as nitrogen, oxygen, hydrogen, noble gases, and some heavier gases like carbon dioxide can be treated as ideal gases within some limits. Generally, a gas behaves more like an ideal gas at a higher temperature and lower pressure, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.

Note: While solving this question, be careful with the units of the terms. All the units of all the terms should be in the same system otherwise the calculated answers can be incorrect.