Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A certain amount of an ideal gas occupies a volume of $1.0m^3$ at a given temperature and pressure. What would be its volume after reducing its pressure to half the initial value and raising the temperature twice the initial value?


Can you answer this question?

1 Answer

0 votes
$P_2 = \large\frac{1}{2}P_1$
$T_2 = 2T_1$
$V_1 = 1.0m^3$
Now $\large\frac{P_1\times1.0}{T_1} = \large\frac{1\times P_1\times V_2}{2\times2\times T_1}$
$\Rightarrow P_1\times1.0 = \large\frac{1\times P_1\times V_2}{4}$
$\Rightarrow 1.0 = \large\frac{1\times V_2}{4}$
$ 4 = V_2$
$V_2 = 4m^3$
Hence answer is (d)
answered Feb 15, 2014 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App