A student forgot to add the reaction mixture | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the volume of the round bottomed flask be $V$.The initial volume of air inside the flask at $T_{1} = 27^{\circ} C = 300 \, K$ is $V_{1} = V$.A

Vedclass · Accepted Answer

$3/5$

Vedclass · Answer

(B) Let the volume of the round bottomed flask be $V$.The initial volume of air inside the flask at $T_{1} = 27^{\circ} C = 300 \, K$ is $V_{1} = V$.At $T_{2} = 477^{\circ} C = 750 \, K$,the volume of air that would occupy the flask is $V_{2}$.According to Charles's law,at constant pressure,$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.Substituting the values,$V_{2} = \frac{V 	imes 750}{300} = 2.5 \, V$.The volume of air expelled is the difference between the final volume and the initial volume: $V_{expelled} = 2.5 \, V - V = 1.5 \, V$.The fraction of air expelled is $\frac{V_{expelled}}{V_{initial}} = \frac{1.5 \, V}{2.5 \, V} = \frac{1.5}{2.5} = \frac{3}{5}$.

Similar Questions

What is the molar volume of an ideal gas at $27\,^oC$ and $1\, atm$ pressure in $L$?

At $25^{\circ} C$ and $760 \,mm \,Hg$ pressure,a gas occupies $600 \,mL$ volume. What will be its pressure at a height where the temperature is $10^{\circ} C$ and the volume of the gas is $640 \,mL$?

What will be the pressure exerted by a mixture of $3.2 \ g$ of methane and $4.4 \ g$ of carbon dioxide contained in a $9 \ dm^3$ flask at $27^{\circ} C$?

$A$ $3.7 \ g$ mass of a gas occupies a certain volume at $25 \ ^\circ C$. The same volume is occupied by $0.184 \ g$ of $H_2$ gas at the same pressure and $17 \ ^\circ C$. Calculate the molar mass of the gas.

$4.4 \ g$ of a gas at $0^{\circ} C$ and $0.82 \ atm$ pressure occupies a volume of $2.73 \ L$. The gas can be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module