Question

The figure shows two vessels with adiabatic walls, one containing 0.1 g of helium (γ = 1.67, M = 4 g mol⁻¹) and the other containing some amount of hydrogen (γ = 1.4, M = 2 g mol⁻¹). Initially, the temperatures of the two gases are equal. The gases are electrically heated for some time during which equal amounts of heat are given to the two gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.

Answer 1

Given:
Mass of He, m_He = 0.1 g
γ₁ = 1.67
Molecular weight of He, M_He = 4 g/mol
M_H2 = ?
M_H2 = 2 g/mol
γ₂ = 1.4

Since it is an adiabatic environment and the system is not dong any external work, the amount of heat given will be used up entirely to raise its internal energy.
For He, dQ = dU = nC_vdT ...(i)
$$\begin{gathered} =\frac{m}{4}\times \frac{\mathrm{R}}{\gamma -1}\times d\mathrm{T} \\ =\frac{0.1}{4}\times \frac{\mathrm{R}}{(1.67-1)}\times d\mathrm{T} \end{gathered}$$
For H₂, dQ =dU = nCvdT ...(ii)
$$\begin{gathered} =\frac{m}{2}\times \frac{\mathrm{R}}{\gamma -1}\times d\mathrm{T} \\ =\frac{m}{2}\times \frac{\mathrm{R}}{1.4-1}\times d\mathrm{T}, \end{gathered}$$

where m is the required mass of H₂.

Since equal amount of heat is given to both gases; so dQ is same in both eq (i) and (ii), we get
$$\begin{gathered} \frac{0.1}{4}\times \frac{R}{0.67}dT \\ =\frac{m}{2}\times \frac{R}{0.4}\times dT \\ \Rightarrow m=\frac{0.1}{2}\times \frac{0.4}{0.67} \\ \Rightarrow m=0.0298\approx 0.03\mathrm{g} \end{gathered}$$