A container is divided into two equal parts A | vedclass

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

Question

(A) Initially,both parts $A$ and $B$ have equal volumes $(V)$ and equal pressures $(P = 1 \ atm)$.When the partition is removed,the total volume becom

Vedclass · Accepted Answer

Pressure remains unchanged in both $A$ and $B$.

Vedclass · Answer

(A) Initially,both parts $A$ and $B$ have equal volumes $(V)$ and equal pressures $(P = 1 \ atm)$.When the partition is removed,the total volume becomes $2V$.According to Boyle's Law,$P_1V_1 = P_2V_2$.For the total system,the final pressure $P_f$ is calculated as: $P_f = \frac{P_A V + P_B V}{2V} = \frac{1 	imes V + 1 	imes V}{2V} = 1 \ atm$.Since the gases mix and occupy the total volume,the partial pressure of each gas decreases to $0.5 \ atm$,but the total pressure of the mixture remains $1 \ atm$ throughout the container.

$A$ container is divided into two equal parts $A$ and $B$ containing $H_2$ and $O_2$ respectively,each at $1 \ atm$ pressure. If the wall separating the two sections is removed,what happens to the pressure?

Similar Questions

What is the density of $SO_2$ gas at $27\,^oC$ temperature and $2\, atm$ pressure in $g/L$? (Atomic weights: $S = 32, O = 16, R = 0.082\, L\, atm\, K^{-1}\, mol^{-1}$)

Explain the mathematical formula,graph,and absolute zero temperature for Charles' Law.

$A$ gas obeys Charles's law in the temperature range $0-500 \ K$ at a given pressure. Its volume changes to zero at a temperature of:

At a temperature of $300 \ K$,if the pressure of $0.09 \ mol$ of a gas is halved,what will be its volume?

If the initial volume of a neon gas is reduced by $15\%$ upon cooling to $75\,^\circ C$,to what temperature in $^\circ C$ should it be heated so that the pressure becomes double?

Vedclass Test Series

Exam Paper Generator

Online Exam Module