Question

if the density of a certain gas at 30 °c and 768 torr is 1.35 kg/m3, its density at stp would be

Answer 1

Ans is (1) 1.48 kg/m3

You can solve it using ideal gas equation

pV=nRT=wMRT

​where p = pressure exerted by the gas
V = volume occupied by that gas
n = no. of moles of the gas
R = Gas constant
T = temperature in Kelvin
w = mass taken of the gas
M = Molar mass of the gas

On rearranging the equation:

pM=wVRTDensity (d) = massVolume=wMThus, pM=dRT by⇒M=dRTp

Given, d = 1.35 kg/m3
p = 768 torr
T = 30 οC = 303 K
R = 62.363 × 10-3 m3 torr K-1 mol-1 (according to units of p and volume)
M = ?
Thus, M = dRTp⇒M = 1.35 kg/m3×62.363×10−3m3 torr K−1mol−1× 303 K768 torr⇒M = 0.0332 kg/mol

Now STP means pressure of 1 bar and temperature of 273 K. Thus the new conditions are:
p = 1 bar
T = 273 K
R = 8.314 × 10-5 bar m3 K-1 mol-1 (again according to units of pressure and volume)
M = 0.0332 kg mol-1 (Molar mass of the gas will not change)
d = ?

Using the same equation, we can find the density at STP as

pM = dRT⇒d=pMRT⇒d=1 bar × 0.0332 kg mol−18.314×10−5bar m3 K−1 mol−1 × 273 K⇒

the answer is
d=1.48 kg/m