Search any question & find its solution
Question:
Answered & Verified by Expert
An ideal gas at temperature $T$, pressure $p$ occupies a volume $V$. If its temperature is halved and pressure doubled, what is its new volume?
Options:
Solution:
1164 Upvotes
Verified Answer
The correct answer is:
$\frac{V}{4}$
Given, initial temperature, pressure and volume of ideal gas,
$T_1=T, p_1=p, V_1=V$
Final temperature, pressure and volume of gas is given as
$T_2=\frac{T_1}{2}=\frac{T}{2}$
$p_2=2 p_1=2 p$
$V_2=?$
According to ideal gas equation,
$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$
$\Rightarrow \quad V_2=\frac{p_1 V_1 T_2}{T_1 p_2}=\frac{p \times V \times T / 2}{T \times 2 p} \Rightarrow V_2=\frac{V}{4}$
$T_1=T, p_1=p, V_1=V$
Final temperature, pressure and volume of gas is given as
$T_2=\frac{T_1}{2}=\frac{T}{2}$
$p_2=2 p_1=2 p$
$V_2=?$
According to ideal gas equation,
$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$
$\Rightarrow \quad V_2=\frac{p_1 V_1 T_2}{T_1 p_2}=\frac{p \times V \times T / 2}{T \times 2 p} \Rightarrow V_2=\frac{V}{4}$
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.