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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^(-1) mol^(-1))`

Answer» Equation for ideal gas,
`PV = (w)/(M) RT`
or `P = (w)/(V) xx (RT)/(M)`
`= d xx (RT)/(M) ((w)/(V) = " density " = d)`
Given, `d = 8.0 kg m^(-3), R = 8.314 K^(-1) mol^(-1)`
`T = - 40 + 273 = 233 K`
and `M = 36.5 g mol^(-1) = 36.5 xx 10^(-3) kg mol^(-1)`
Substituting the values in the above equation,
`P = (8.0 xx 8.314 xx 233)/(36.5 xx 10^(-3)) = 424.58 xx 01^(3) Pa`


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