Search any question & find its solution
Question:
Answered & Verified by Expert
1 mole of $\mathrm{H}_2$ gas is contained in a box of volume $V=1.00 \mathrm{~m}^3$ at $T=300 \mathrm{~K}$. The gas is heated to a temperature of $T=3000 \mathrm{~K}$ and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)
Options:
Solution:
1867 Upvotes
Verified Answer
The correct answer is:
20 times the pressure initially
According to gas equation
$$
P V=n R T
$$
or $\quad \frac{P_2 V_2}{T_2}=\frac{P_1 V_1}{T_1} \quad$ or $\quad \frac{P_2}{P_1}=\frac{V_1}{V_2} \cdot \frac{T_2}{T_1}$
Here, $T_2=3000 \mathrm{~K}, T_1=300 \mathrm{~K}$
Since $\mathrm{H}_2$ splits into hydrogen atoms, therefore volume become half i.e., $V_2=\frac{1}{2} V_1$
$$
\begin{aligned}
& \therefore \quad \frac{P_2}{P_1}=\frac{V_1}{\frac{1}{2} V_1} \times \frac{3000}{300} \\
& \text { or } \frac{P_2}{P_1}=20
\end{aligned}
$$
$$
P V=n R T
$$
or $\quad \frac{P_2 V_2}{T_2}=\frac{P_1 V_1}{T_1} \quad$ or $\quad \frac{P_2}{P_1}=\frac{V_1}{V_2} \cdot \frac{T_2}{T_1}$
Here, $T_2=3000 \mathrm{~K}, T_1=300 \mathrm{~K}$
Since $\mathrm{H}_2$ splits into hydrogen atoms, therefore volume become half i.e., $V_2=\frac{1}{2} V_1$
$$
\begin{aligned}
& \therefore \quad \frac{P_2}{P_1}=\frac{V_1}{\frac{1}{2} V_1} \times \frac{3000}{300} \\
& \text { or } \frac{P_2}{P_1}=20
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.