Answer :

Given:


Two vessels with adiabatic walls, one contains 0.1g of helium (γ = 1.67, M = 4 g mol–1) and the other contains some amount of hydrogen (γ= 1.4, M = 2g mol–1)


The gasses are given the same amount of heat.


The temperature rises through the same amount.


0.1g of helium = 0.1/4 mole = 0.025mole


Let there be n moles of hydrogen in the other vessel.


= and so,


As the vessels are of constant volume there will be no work done by the gasses. The heat supplied will totally be used to increase internal energy.


Therefore, where Q is the heat supplied, n is the number of moles, Cv is the specific heat capacity of gas at constant volume, T is the change in temperature.


For helium,


For hydrogen, we assume for both cases the rise of temperature is T.


As per question,




n = 0.015


Again, Molar mass of hydrogen = 2g mol–1


Therefore, 0.015 mole of hydrogen hydrogen


Thus, there is 0.03g of hydrogen in the vessel.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Two vessels A andHC Verma - Concepts of Physics Part 2

4.0 g of helium oHC Verma - Concepts of Physics Part 2

Figure shows two HC Verma - Concepts of Physics Part 2

An amount Q of heHC Verma - Concepts of Physics Part 2

The volume of an HC Verma - Concepts of Physics Part 2

A gas is enclosedHC Verma - Concepts of Physics Part 2

Two samples A andHC Verma - Concepts of Physics Part 2

A sample of an idHC Verma - Concepts of Physics Part 2

An ideal gas (C<sHC Verma - Concepts of Physics Part 2

Half mole of an iHC Verma - Concepts of Physics Part 2