The Macroscopic Picture
Imagine a sturdy chamber filled with an ideal gas
We are given its macroscopic properties: a volume of V=2 m3, and a pressure of p=3×106 Pa. These two values are the only keys we need to unlock the total energy hidden within the chaotic motion of the gas molecules.
The Master Equation
Now, what exactly is the "energy" of this gas? For an ideal gas, unless specified otherwise, we are talking about its translational kinetic energy
The gas molecules are zipping around in three-dimensional space, giving them f=3 degrees of freedom.
The fundamental formula for the internal energy of an ideal gas is:
E=2fnRT
Since the degree of freedom f is 3, we get E=23nRT. But wait, we don't know the temperature T or the number of moles n! Here is the catch—we can use the ideal gas law, pV=nRT, to elegantly replace nRT with pV.
So, our master equation transforms into a purely macroscopic form:
E=23pV
The Final Calculation
Let's bring back those values we noted earlier
We need to carefully substitute the pressure and the volume into our master equation. Don't rush through this; let's set it up properly.
Look closely at the equation... do you see the magic? The 2 in the denominator perfectly cancels out the 2 from the volume. It's a very simple calculation now. We are simply left with 3 multiplied by 3×106.
Our final answer is 9×106 J, which perfectly matches option (c).
A Quick Thought Experiment: What if the question explicitly mentioned a diatomic gas and asked for the total internal energy? Then we would use f=5 degrees of freedom, and the formula would be E=25pV. Always watch out for these subtle hints in JEE questions!