Calculate the volume occupied by 8-8 g of CO2 at 31-1-C and 1 bar pressure- Given- R-0-083 bar L K-1 mol-1

Number of moles of CO_2(n)=Mass of CO_2Molar mass =8.8 g44 g mol^ 1=0.2 mol Pressure of CO_2(P)=1 bar R=0.083 bar L K^ 1 mol^ 1 Temprature (T)=273+31.1 =304. ...

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Standard XII

Chemistry

First Law of Thermodynamics

Calculate the...

Question

Calculate the volume occupied by $8.8 g$ of $C O_{2}$ at ${31.1}^{\circ} C$ and $1 bar$ pressure. (Given, $R = 0.083 {bar L K}^{- 1} {mol}^{- 1}$ )

5.048 L

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4.048 L

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Solution

The correct option is A $5.048 L$
Number of moles of $C O_{2} (n) = \frac{Mass of C O_{2}}{Molar mass}$
$= \frac{8.8 g}{44 {g mol}^{- 1}} = 0.2 mol$
Pressure of $C O_{2} (P) = 1 bar$
$R = 0.083 bar {L K}^{- 1} {mol}^{- 1}$
Temprature (T) $= 273 + 31.1$
$= 304.1 K$
Sinec from gas equation $P V = n R T$
$V = \frac{n R T}{P} = \frac{0.2 \times 0.083 \times 304.1}{1 b a r}$
$= 5.048 L$

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Calculate the volume occupied by 8.8 g of CO2 at 31.1∘C and 1 bar pressure. (Given, R=0.083 bar L K−1 mol−1)

The correct option is A 5.048 LNumber of moles of CO2(n)=Mass of CO2Molar mass =8.8 g44 g mol−1=0.2 mol Pressure of CO2(P)=1 bar R=0.083 bar L K−1 mol−1 Temprature (T)=273+31.1 =304.1K Sinec from gas equation PV=nRT V=nRTP=0.2×0.083×304.11bar =5.048 L

Calculate the volume occupied by $8.8 g$ of $C O_{2}$ at ${31.1}^{\circ} C$ and $1 bar$ pressure. (Given, $R = 0.083 {bar L K}^{- 1} {mol}^{- 1}$ )