Ncert Exercises & Solutions

\(1\) mole of \(\mathrm{H}_{2}\) gas is contained in a box of volume \(V = 1.00 ~\text m^{3}\) at \(T = 300 ~\text{K}\). The gas is heated to a temperature of \(T = 3000~\text{K}\) and the gas gets converted to a gas of hydrogen atoms. The final pressure would be:
(considering all gases to be ideal)

1.	same as the pressure initially.
2.	\(2\) times the pressure initially.
3.	\(10\) times the pressure initially.
4.	\(20\) times the pressure initially.

Hint: \(PV=nRT\)

Step 1: Find the final number of moles.

Consider the diagram, when the molecules break into atoms, the number of moles would become twice.

As gas breaks, the number of moles becomes twice of initial, so \(𝑛 _2 = 2 𝑛 _1 .\)

Step 2: Apply the ideal gas equation.
Now, by the ideal gas equation, \(PV =nRT\)
\(P \propto n T\)
The number of moles becomes twice of initial, so \(𝑛 _2 = 2 𝑛 _1 .\)
\(\frac{P_{1}}{P_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}} \Rightarrow \frac{{1} \times 300}{{2} \times 3000}\)
\(P_{2} = 20 P_{1}\)
Thus, the final pressure becomes \(20\) times the initial pressure.
Hence, option (4) is the correct answer.

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