Question

For the given isotherm for one mole of an ideal gas, which follows Boyle's law, what will be the value of temperature.

• A. $8.2\text{K}$
• B. $122\text{K}$
• C. $947\text{K}$
• D. $48\text{K}$

Answer 1

The correct option is B $122\text{K}$


According to Ideal gas law,
$PV=\text{nRT}$
According to this graph:
$y=mx+c$
${\text{log}}_{10}P+{\text{log}}_{10}V=2$
(c = $lognRT=2$)

${\text{log}}_{10}\left(PV\right)=2\Rightarrow PV={10}^2$
$\Rightarrow PV=100$
Now, $PV=nRT$
$100=nRT$
For 1 mol
$100=RT$
$T=\frac{100}{\left(0.0820\right)}=1219.5\approx 1220\text{K}$