Two vessels with adiabatic walls are shown in the figure. One vessel holds 0.1 g of helium ( γ = 1.67 , M = 4 g mol^ -1 ) while the other holds an unknown amount of hydrogen ( γ = 1.4 , M = 2 g mol^ -1 ). The initial

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Quantrex Academy · Accepted Answer

Correct answer: A. 0.03 g

Since the gases are heated in closed vessels, the process is isochoric (constant volume).

The heat supplied at constant volume is given by:

Q = n C_v \Delta T

Given that equal amounts of heat are supplied (

Q_{He} = Q_{H_2}

) and the temperature increase is the same (

\Delta T_{He} = \Delta T_{H_2}

), we have:

n_{He} C_{v,He} = n_{H_2} C_{v,H_2}

We know that

C_v = \dfrac{R}{\gamma - 1}

and

n = \dfrac{m}{M}

.

For Helium (monoatomic):

m_{He} = 0.1\text{ g}

,

M_{He} = 4\text{ g mol}^{-1}

,

\gamma_{He} = 1.67 \approx \dfrac{5}{3}

C_{v,He} = \dfrac{R}{\dfrac{5}{3} - 1} = \dfrac{3}{2}R

For Hydrogen (diatomic):

M_{H_2} = 2\text{ g mol}^{-1}

,

\gamma_{H_2} = 1.4 = \dfrac{7}{5}

C_{v,H_2} = \dfrac{R}{\dfrac{7}{5} - 1} = \dfrac{5}{2}R

Substituting these values into the equation:

\dfrac{0.1}{4} \times \dfrac{3}{2}R = \dfrac{m_{H_2}}{2} \times \dfrac{5}{2}R

\dfrac{0.3}{8} = \dfrac{5 m_{H_2}}{4}

m_{H_2} = \dfrac{0.3 \times 4}{8 \times 5}

m_{H_2} = \dfrac{1.2}{40} = 0.03\text{ g}

Two vessels with adiabatic walls are shown in the figure. One...

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