Search any question & find its solution
Question:
Answered & Verified by Expert
A $10 \mathrm{~g}$ of a gas at atmospheric pressure is cooled from $273^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ keeping the volume constant, its pressure would become
Options:
Solution:
2501 Upvotes
Verified Answer
The correct answer is:
$1 / 2$ atm
$T_1=273^{\circ} \mathrm{C}=273+273^{\circ} \mathrm{K}=546^{\circ} \mathrm{K}$
$T_2=0^{\circ} \mathrm{C}=273+0^{\circ} \mathrm{C}=273^{\circ} \mathrm{K}$
$P_1=1 . P_2=$ ?
According to Gay-Lussac's law
$\frac{P_1}{T_1}=\frac{P_2}{T_2} \quad \therefore \quad P_2=\frac{P_1 T_2}{T_1}=\frac{1 \times 273^{\circ} \mathrm{K}}{546^{\circ} \mathrm{K}} \quad$ atm; $\quad \frac{1}{2} \mathrm{~atm}$.
$T_2=0^{\circ} \mathrm{C}=273+0^{\circ} \mathrm{C}=273^{\circ} \mathrm{K}$
$P_1=1 . P_2=$ ?
According to Gay-Lussac's law
$\frac{P_1}{T_1}=\frac{P_2}{T_2} \quad \therefore \quad P_2=\frac{P_1 T_2}{T_1}=\frac{1 \times 273^{\circ} \mathrm{K}}{546^{\circ} \mathrm{K}} \quad$ atm; $\quad \frac{1}{2} \mathrm{~atm}$.
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.