If two moles of an ideal gas at $546 \mathrm{~K}$ occupy a volume of $44.8 \mathrm{~L}$. What is the…
If two moles of an ideal gas at $546 \mathrm{~K}$ occupy a volume of $44.8 \mathrm{~L}$. What is the pressure of ideal gas at $546 \mathrm{~K} ?\left(\mathrm{R}=0 \cdot 0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
20 atm
$0 \cdot 2$ atm
$0 \cdot 5$ atm
$2 \cdot 0$ atm
Solution
$\mathrm{n}=2 \mathrm{~mol}, \mathrm{~T}=546 \mathrm{~K}, \mathrm{~V}=44.8 \mathrm{~L}, \mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \mathrm{P}=?$
According to ideal gas equation,
$\mathrm{PV}=\mathrm{n} \mathrm{R} \mathrm{T}$
$\therefore P=\frac{n R T}{V}=\frac{2 \times 0.0821 \times 546}{44.8}=2.0 \mathrm{~atm}$