Thermal Physics> A vessel of...

A vessel of volume 30 l is seprated into three equal parts by stationary semipermeable thin member as shown in the fig.the left middle and right parts are filled with hydrogen =30gm mass of oxygen is 160 gm and of nitrogen is 70 . the left partition lets through only hydrogen ,while the right partition lets through hydrogen and nitrogen. what will be the presssure in each part of vessel after the equilibrium has been set in if the vessel is kept at constant temp T=300k
fig

Navjyot Kalra , 12 Years ago

Grade 10

1 Answers

Askiitians Tutor Team

To determine the pressure in each part of the vessel after equilibrium has been established, we need to analyze the behavior of the gases involved and apply the principles of gas laws, particularly Dalton's Law of Partial Pressures and the Ideal Gas Law. Let's break this down step by step.

Understanding the Setup

We have a vessel divided into three equal parts, each with a volume of 10 liters (since the total volume is 30 liters). The gases present are:

Left part: Hydrogen (H₂) - 30 g
Middle part: Oxygen (O₂) - 160 g
Right part: Nitrogen (N₂) - 70 g

Calculating Moles of Each Gas

First, we need to convert the masses of the gases into moles using the molar masses:

Molar mass of H₂ = 2 g/mol
Molar mass of O₂ = 32 g/mol
Molar mass of N₂ = 28 g/mol

Now, we can calculate the number of moles:

For H₂: n_H2 = 30 g / 2 g/mol = 15 moles
For O₂: n_O2 = 160 g / 32 g/mol = 5 moles
For N₂: n_N2 = 70 g / 28 g/mol = 2.5 moles

Behavior of Gases and Equilibrium

In this setup, the left partition allows only hydrogen to pass through, while the right partition allows both hydrogen and nitrogen. At equilibrium, the gases will distribute themselves according to their partial pressures.

Applying Dalton's Law

Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each gas. The partial pressure of a gas can be calculated using the Ideal Gas Law:

P = (nRT) / V

Where:

P = pressure
n = number of moles
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature in Kelvin (300 K)
V = volume in liters

Calculating Pressures in Each Part

Now, let's calculate the pressures in each section:

Left Part (Hydrogen Only)

For the left part, we have:

P_H2 = (n_H2RT) / V = (15 moles * 0.0821 L·atm/(K·mol) * 300 K) / 10 L

Calculating this gives:

P_H2 = (15 * 0.0821 * 300) / 10 = 36.945 atm

Middle Part (Oxygen Only)

For the middle part, we only have oxygen:

P_O2 = (n_O2RT) / V = (5 moles * 0.0821 * 300) / 10

Calculating this gives:

P_O2 = (5 * 0.0821 * 300) / 10 = 12.315 atm

Right Part (Hydrogen and Nitrogen)

In the right part, both hydrogen and nitrogen are present. We need to calculate their partial pressures:

For hydrogen in the right part:

P_H2 = (n_H2RT) / V = (15 moles * 0.0821 * 300) / 10 = 36.945 atm

For nitrogen:

P_N2 = (n_N2RT) / V = (2.5 moles * 0.0821 * 300) / 10

Calculating this gives:

P_N2 = (2.5 * 0.0821 * 300) / 10 = 6.1575 atm

Now, summing the partial pressures for the right part:

P_Total = P_H2 + P_N2 = 36.945 + 6.1575 = 43.1025 atm

Final Pressures in Each Part

After equilibrium is reached, the pressures in each part of the vessel will be:

Left Part (Hydrogen): 36.945 atm
Middle Part (Oxygen): 12.315 atm
Right Part (Hydrogen and Nitrogen): 43.1025 atm

This analysis shows how the distribution of gases and their respective properties affect the final pressures in a closed system. Understanding these principles is crucial in fields such as chemistry and engineering, where gas behavior plays a significant role in various applications.