If 2 moles of an ideal gas at $546 \mathrm{~K}$ has volume of $44 \cdot 8 \mathrm{~L}$, then what will be…
If 2 moles of an ideal gas at $546 \mathrm{~K}$ has volume of $44 \cdot 8 \mathrm{~L}$, then what will be it's pressure ? $(\mathrm{R}=0.082)$
- 1.998atm
- 1.098atm
- 2.408atm
- 3.129atm
Solution
(D) $\mathrm{n}=2$ moles $, \mathrm{T}=546 \mathrm{~K}, \quad \mathrm{~V}=44.8 \mathrm{~L}$, $\mathrm{R}=0.082, \quad \mathrm{P}=?$
According to ideal gas equation, $\mathrm{PV}=\mathrm{nRT} \quad \therefore \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$ $\therefore \mathrm{P}=\frac{2 \times 0.082 \times 546}{44.8}=1.998 \mathrm{~atm}$
Asked in: MHT CET 2020 (16 Oct Shift 2)
Practice more States of Matter questions on Aicharya