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A gas is present at a pressure of $2 \mathrm{~atm}$. What should be the increase in pressure, so that the volume of the gas can be decreased to $\frac{1}{4}$ th of the initial volume at constant 4 temperature?
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$8 \mathrm{~atm}$
(d) Given, initial pressure $=2 \mathrm{~atm}$
Final volume $=\frac{1}{4}$ th of initial volume $(V)$
Let, initial volume $=V$
then, final volume $=\frac{V}{4}($ at constant temperature $)$
$\because$ Final pressure $=p_2$ (to find $)$
$V_1 p_1=V_2 p_2$
$\therefore \quad p_2=\frac{V_1 p_1}{V_2}=\frac{V}{\frac{V}{4}} \times 2=8 \mathrm{~atm}$
Hence, increase in pressure (i.e. final pressure) $=8 \mathrm{~atm}$
Final volume $=\frac{1}{4}$ th of initial volume $(V)$
Let, initial volume $=V$
then, final volume $=\frac{V}{4}($ at constant temperature $)$
$\because$ Final pressure $=p_2$ (to find $)$
$V_1 p_1=V_2 p_2$
$\therefore \quad p_2=\frac{V_1 p_1}{V_2}=\frac{V}{\frac{V}{4}} \times 2=8 \mathrm{~atm}$
Hence, increase in pressure (i.e. final pressure) $=8 \mathrm{~atm}$
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