Question

Density of a gas is found to be 5.46g/dm³ at 27°C at 2 bar pressure. What will be it's density at STP ?

➙ Need explained answer without copying !
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Answer 1

Answer:

Density of the gas at STP is 3.0 $g/dm^{3}$

Explanation:

According to ideal gas equation , PV = n RT

where, P is the pressure of the gas

           V is the volume of the gas

           n is the number of moles of the gas

           R is the universal gas constant

           T is the temperature

we can also write the ideal gas equation in terms of density(d)

i.e., $d=\frac{PM}{RT}$   , where M is the molar mass of gas

Here, d1 = 5.46 $g/dm^{3}$

         P1 = 2 bar

         T1 = 273+27=300 K

At STP condition

        P2 = 1 bar

        T2 = 273 K

        d2 = ?

$d1=\frac{P1M}{RT1}$      ---(1)

$d2=\frac{P2M}{RT2}$      ---(2)

(1) ÷(2) ⇒ $\frac{d1}{d2} = \frac{P1T2}{T1P2}$

i.e., $d2 = \frac{P2T1d1}{P1 T2}$

          $=\frac{1*300*5.46}{2*273} = 3.0 g/dm^{3}$

∴ density of the gas at STP is 3.0 $g/dm^{3}$

         

Answer 2

Answer:

Density of the gas at STP is 3.0 g/dm^{3}g/dm

3

Explanation:

According to ideal gas equation , PV = n RT

where, P is the pressure of the gas

V is the volume of the gas

n is the number of moles of the gas

R is the universal gas constant

T is the temperature

we can also write the ideal gas equation in terms of density(d)

i.e., d=\frac{PM}{RT}d=

RT

PM

, where M is the molar mass of gas

Here, d1 = 5.46 g/dm^{3}g/dm

3

P1 = 2 bar

T1 = 273+27=300 K

At STP condition

P2 = 1 bar

T2 = 273 K

d2 = ?

d1=\frac{P1M}{RT1}d1=

RT1

P1M

---(1)

d2=\frac{P2M}{RT2}d2=

RT2

P2M

---(2)

(1) ÷(2) ⇒ \frac{d1}{d2} = \frac{P1T2}{T1P2}

d2

d1

=

T1P2

P1T2

i.e., d2 = \frac{P2T1d1}{P1 T2}d2=

P1T2

P2T1d1

=\frac{1*300*5.46}{2*273} = 3.0 g/dm^{3}=

2∗273

1∗300∗5.46

=3.0g/dm

3

∴ density of the gas at STP is 3.0 g/dm^{3}g/dm

3