Question

Density of a gas is formed to be 5.46g/dm3 at 27 degree at 2 bar pressure what will be it's density at STP.

Answer 1

Data:
d' (initial density) = 5.46 g/dm³
T' (initial temperature) = 27º C 
Converting to Kelvin: TK = TC + 273 → TK = 27 + 273 = 300 K
P' (Initial pressure) = 2 bar
d" (final density) = ? (in STP)
P" (Final pressure) = 1 bar (Standard Temperature and Pressure)
T" (final temperature) = 273 K (Standard Temperature and Pressure)

We have the following equation of an ideal gas (Clapeyron equation)
$P*V = n*R*T$
If: 
$n = \frac{m}{MM}$ m(mass) and mm (Molar Mass)
So: 
$P*V = \frac{m}{MM} *R*T$
Product of extremes equals product of means:
$\frac{P*V}{R*T} = \frac{m}{MM}$
If: MM (Molar Mass) = V (Molar Volume)
So: 
$\frac{P*V}{R*T} = \frac{m}{V}$

Knowing that the density formula is $d = \frac{m}{V}$, we have:
$\frac{P*V}{R*T} = d$

Soon: $d = \frac{P*V}{R*T}$

Adopting: $d = \frac{d'}{d"}$ and $P*V = P'*T''$ and $R*T = P''*T''$

We have:
$\frac{d'}{d''} = \frac{P'*T''}{P''*T'}$

Solving:
$\frac{d'}{d''} = \frac{P'*T''}{P''*T'}$
$\frac{5.46}{d''} = \frac{2*273}{1*300}$
$\frac{5.46}{d''} = \frac{546}{300}$
Product of extremes equals product of means:
$546*d'' = 5.46*300$
$546d'' = 1638$
$d'' = \frac{1638}{546}$
$d'' = 3\:g/cm^3\stackrel{(in\:STP)}{\longrightarrow} \boxed{\boxed{d'' = 3\:g/cm^{-3}}}\end{array}}\qquad\quad\checkmark$