Question

Density of a gas is found to be 5.46 g/$dm^{3}$ at 27°C at 2 bar pressure. What will be its density at STP ?

Answer 1

For an ideal gas,

$Density\quad (\rho )\quad =\quad P\quad \times \quad \frac { M }{ RT }$

From the given data,

Density = 5.46

Pressure = 2 atm

Temperature = 300 K

$5.46\quad =\quad 2\quad \times \quad \frac { M }{ 300\quad \times \quad R } \quad$ ............(1)

At STP,

${ \rho }_{ STP }\quad =\quad 1\quad \times \quad \frac { M }{ 300\quad \times \quad R } \quad$..........(2)

From (1) and (2)

We get,

${ \rho}_{ STP }\quad =\quad \left( \frac { 1\quad \times \quad M }{ 273\quad \times \quad R }\right) \quad \times \left( \frac { 300\quad \times \quad R }{ 2\quad \times \quad M }\right)$

$=\quad \frac { 300}{ 273\quad \times \quad 2 } \quad$

${ \rho}_{ STP }\quad =\quad 3.00\quad g/\quad d{ m }^{ 3 }$

Therefore, the density of the gas at STP will be $3\quad g/d{ m }^{ 3 }$

Answer 2

For an ideal gas,

Density\quad (\rho )\quad =\quad P\quad \times \quad \frac { M }{ RT }Density(ρ)=P×RTM​

From the given data,

Density = 5.46

Pressure = 2 atm

Temperature = 300 K

5.46\quad =\quad 2\quad \times \quad \frac { M }{ 300\quad \times \quad R } \quad5.46=2×300×RM​ ............(1)

At STP,

{ \rho }_{ STP }\quad =\quad 1\quad \times \quad \frac { M }{ 300\quad \times \quad R } \quadρSTP​=1×300×RM​ ..........(2)

From (1) and (2)

We get,

{ \rho}_{ STP }\quad =\quad \left( \frac { 1\quad \times \quad M }{ 273\quad \times \quad R }\right) \quad \times \left( \frac { 300\quad \times \quad R }{ 2\quad \times \quad M }\right)ρSTP​=(273×R1×M​)×(2×M300×R​)

=\quad \frac { 300}{ 273\quad \times \quad 2 } \quad=273×2300​

{ \rho}_{ STP }\quad =\quad 3.00\quad g/\quad d{ m }^{ 3 }ρSTP​=3.00g/dm3

Therefore, the density of the gas at STP will be 3\quad g/d{ m }^{ 3 }3g/dm3