Search any question & find its solution
Question:
Answered & Verified by Expert
If pressure of real gas $\mathrm{O}_{2}$ in a container is given by $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$, then the mass of the gas in the container is
Options:
Solution:
1981 Upvotes
Verified Answer
The correct answer is:
$16 \mathrm{gm}$
$\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}$
$2\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\mathrm{RT}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\frac{\mathrm{RT}}{2}$
$\therefore \mathrm{n}=1 / 2 \mathrm{~mole}$
$\therefore$ Mass of $\mathrm{O}_{2}=\frac{1}{2} \times 32=16 \mathrm{gm}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}$
$2\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\mathrm{RT}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\frac{\mathrm{RT}}{2}$
$\therefore \mathrm{n}=1 / 2 \mathrm{~mole}$
$\therefore$ Mass of $\mathrm{O}_{2}=\frac{1}{2} \times 32=16 \mathrm{gm}$
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.