The density of a gas is found to be 2.07 ${\mathrm{gL}}^{-1}$at 30°C and 2 atm pressure. What is its density at NTP

The given data is as follows:

${\mathrm{d}}_1=2.07 \mathrm{g}/\mathrm{lit}$

${\mathrm{T}}_1=30 + 273=303 \mathrm{k}; {\mathrm{T}}_{\mathrm{k}}=°\mathrm{C}+273$

${\mathrm{T}}_2=273\mathrm{K}; \mathrm{NTP} \mathrm{means} \mathrm{normal} \mathrm{Temp} \mathrm{and} \mathrm{pressure}$${\mathrm{P}}_1=2 \mathrm{atm} {\mathrm{P}}_2=1 \mathrm{atm}$

Then   ${\mathrm{d}}_2=?$

From ideal gas equation,

$\mathrm{PV}=\frac{\mathrm{W}}{\mathrm{G}.\mathrm{M}.\mathrm{W}}\times \mathrm{RT}; \mathrm{since}, \mathrm{n} =\frac{\mathrm{weight} }{\mathrm{gram} \mathrm{molecular} \mathrm{weight}}$$\mathrm{Then} \mathrm{P}\times \mathrm{GMW}=\mathrm{dRT}$

Consider the equation for same gas at two different conditions

(G.M.W is same)

$\frac{{\mathrm{d}}_1{\mathrm{T}}_1}{{\mathrm{P}}_1}=\frac{d_2{T}_2}{P_2}$

$\Rightarrow \frac{2.07\times 303}{2}=\frac{{\mathrm{d}}_2\times 273}{1}$$d_2=\frac{2.07\times 303}{2\times 273}=1.149g/litr$$\mathrm{Density} \mathrm{of} \mathrm{gas} \mathrm{at} \mathrm{NTP}\hspace{0.17em}\mathrm{is} 1.149g/litr$