What is the pressure of HCl gas at -40-C if its density is 8-0 kg m-3- R - 8-314 J K-1 mol-1-

For ideal gas, PV = wMRT or P = wV×RTM = d×RTM (wV = density = d ) Given, d = 8.0 kg m^ 3, R = 8.314 J K^ 1 mol^ 1 T = 40 + 273 = 233 K ...

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Standard XII

Physics

Boiling

What is the p...

Question

What is the pressure of $H C l$ gas at $- 40^{\circ} C$ if its density is $8.0 k g m^{- 3}$ ? $(R = 8.314 J K^{- 1} m o l^{- 1})$ .

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Solution

For ideal gas,
$P V = \frac{w}{M} R T$
or $P = \frac{w}{V} \times \frac{R T}{M}$
$= d \times \frac{R T}{M} (\frac{w}{V} = d e n s i t y = d)$

Given, $d = 8.0 k g m^{- 3}, R = 8.314 J K^{- 1} m o l^{- 1}$
$T = - 40 + 273 = 233 K$
and $M = 36.5 g m o l^{- 1} = 36.5 \times 10^{- 3} k g m o l^{- 1}$
Substituting the values in the above equation we get,

$P = \frac{8.0 \times 8.314 \times 233}{36.5 \times 10^{- 3}} = 424.58 \times 10^{3} P a s c a l$

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What is the pressure of HCl gas at −40∘C if its density is 8.0 kg m−3? (R=8.314 J K−1mol−1).

For ideal gas,PV=wMRTor P=wV×RTM=d×RTM(wV=density=d)Given, d=8.0 kg m−3,R=8.314 J K−1mol−1T=−40+273=233 Kand M=36.5 g mol−1=36.5×10−3kg mol−1Substituting the values in the above equation we get, P=8.0×8.314×23336.5×10−3=424.58×103Pascal

What is the pressure of $H C l$ gas at $- 40^{\circ} C$ if its density is $8.0 k g m^{- 3}$ ? $(R = 8.314 J K^{- 1} m o l^{- 1})$ .