Three bulbs are filled with CH_4,CO_2,and Ne | vedclass

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

Question

(B) Using the ideal gas equation $PV = nRT$,since $T$ is constant,$n = \frac{PV}{RT}$.For $CH_4$: $n_1 = \frac{2 	imes 2}{RT} = \frac{4}{RT}$.For $CO

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Using the ideal gas equation $PV = nRT$,since $T$ is constant,$n = \frac{PV}{RT}$.For $CH_4$: $n_1 = \frac{2 	imes 2}{RT} = \frac{4}{RT}$.For $CO_2$: $n_2 = \frac{4 	imes 3}{RT} = \frac{12}{RT}$.For $Ne$: $n_3 = \frac{3 	imes 4}{RT} = \frac{12}{RT}$.Total moles $n_T = n_1 + n_2 + n_3 = \frac{4 + 12 + 12}{RT} = \frac{28}{RT}$.Total volume $V_T = 2 + 3 + 4 = 9 \ L$.Final pressure $P_T = \frac{n_T RT}{V_T} = \frac{28}{RT} 	imes \frac{RT}{9} = \frac{28}{9} \approx 3.11 \ atm$.The nearest integer is $3$.

Three bulbs are filled with $CH_4$,$CO_2$,and $Ne$ as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout,the pressure of the system is found to be $...... \ atm$. (Nearest integer)

Similar Questions

In the equation $PV = nRT$,the number of moles per liter can be expressed as:

Under what conditions does a gas behave ideally? Why?

The correct combined gas law equation is:

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

When a graph of the product of pressure and volume $(PV)$ versus pressure $(P)$ is plotted for a fixed amount of gas at a constant temperature,what type of line is obtained?

Vedclass Test Series

Exam Paper Generator

Online Exam Module