Question

The density of a gas at $80cm$ Hg and ${50}^{o}C$ is $1.85gm{L}^{-1}$. What would be its density at NTP?

• A. $3.07g/ml$
• B. $2.07g/ml$
• C. $1.56g/ml$
• D. $4.5g/ml$

Answer 1

The correct option is C $1.56g/ml$

From ideal gas equation-

$PV=nRT$

whereas,

$P=$ pressure of the gas

$V=$ volume it occupies

$n=$ number of moles of gas present in the sample $=\frac{\text{mass of gas}\left(m\right)}{\text{molar mass of gas}\left(M\right)}$

$R=$ universal gas constant $=0.0821atmL{mol}^{-1}{K}^{-1}$

$T=$ absolute temperature of the gas

$\therefore PV=\frac{m}{M}RT$

$\Rightarrow PM=\frac{m}{V}RT$

As we know that,

$\frac{m}{V}=\rho \left(\text{density}\right)$

$\therefore PM=\rho RT$

For same gas,

Density of a gas is directly proportional to the pressure, i.e.,

$\rho \propto \frac{P}{T}$

$\therefore \frac{\rho }{{\rho }^{\prime }}=\frac{P}{P^{\prime }}\times \frac{T^{\prime }}{T}$

$\Rightarrow {\rho }^{\prime }=\frac{\rho \times P^{\prime }\times T}{P\times T^{\prime }}$

Given that:-

$P=80cmHg=\frac{80}{76}atm=1.0526$

$\rho =1.85g{L}^{-1}$

$T=50℃=(50+273)K=323K$

At NTP,

$P^{\prime }=1atm$

$T^{\prime }=20℃=(20+273)K=293K$

${\rho }^{\prime }=?$

$\therefore {\rho }^{\prime }=\frac{1.85\times 1\times 273}{1.0526\times 293}=1.637g{L}^{-1}\approx 1.56g{L}^{-1}$

Hence, density at NTP will be $1.56g{L}^{-1}$ .