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Question: Density of a gas is found to be \(5.46g/dm^{3}\)at \(27^{0}C\)and 2 bar pressure. What will be its ...

Density of a gas is found to be 5.46g/dm35.46g/dm^{3}at

270C27^{0}Cand 2 bar pressure. What will be its density at STP?

A

3.0gdm−33.0gdm^{- 3}

B

5.0gdm−35.0gdm^{- 3}

C

6.0gdm−36.0gdm^{- 3}

D

10.82gdm−310.82gdm^{- 3}

Answer

3.0gdm−33.0gdm^{- 3}

Explanation

Solution

: P=nvRT=mv×RTM=dRTMP = \frac{n}{v}RT = \frac{m}{v} \times \frac{RT}{M} = \frac{dRT}{M}

Or d = PMRTord∝PT\frac{PM}{RT}ord \propto \frac{P}{T}

d1d2=P1P2×T2T1ord2=d1×P2×T1P1×T2\frac{d_{1}}{d_{2}} = \frac{P_{1}}{P_{2}} \times \frac{T_{2}}{T_{1}}ord_{2} = \frac{d_{1} \times P_{2} \times T_{1}}{P_{1} \times T_{2}}

d2=5.46×1×3002×273d_{2} = \frac{5.46 \times 1 \times 300}{2 \times 273}=3.0gdm−33.0gdm^{- 3}