Question

At standard temperature and pressure (0 ∘C∘C and 1.00 atmatm ), 1.00 molmol of an ideal gas occupies a volume of 22.4 LL. What volume would the same amount of gas occupy at the same pressure and 25 ∘C∘C ?

Answer 1

Answer:

Final volume will be 24.45 L

Explanation:

We have given initial temperature $T_1=0^{\circ}C=0+273=273K$

Pressure is $P_1=1atm$

Volume occupied $V_1=22.4lL$

From ideal gas equation $PV=nRT$

$\frac{PV}{T}=constant$

So $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Final temperature $T_2=25+273=298K$

Pressure is remain constant so $P_2=1atm$

We have to fond the volume $V_2$

So $\frac{1\times 22.4}{273}=\frac{1\times V_2}{298}$

$V_2=24.45L$

So final volume will be 24.45 L