Q. 28

# Figure shows two

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.

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