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.
Figure

Answer 1

The mass of Hydrogen is 0.03 gram

Explanation:

Given Data

Mass of Helium, $m_H_e = 0.1 g$

$\gamma_1 = 1.67$

Molecular weight of Helium, $M_H_e = 4 g/mol$

MH2 = 2 g/mol

$\gamma_2 = 1.4$

To find the mass of Hydrogen (m)

Since it is an adiabatic environment and no external work is done by the system, the amount of heat provided will be used up entirely to elevate its internal energy.  

For Helium (He), dQ = dU = nCvdT   ...(i)

$=\frac{m}{4} \times \frac{R}{\gamma-1} \times d T$

$=\frac{0.1}{4} \times \frac{R}{1.67-1} \times d T$

in case of  Hydrogen (H2), dQ =dU = nCvdT   ...(ii)

$=\frac{m}{2} \times \frac{R}{\gamma-1} \times d T$

$=\frac{m}{2} \times \frac{R}{1.4-1} \times d T$

Where m is the H2 mass required.  

Since both gasses receive the same amount of heat; thus, dQ is the same in both equation (i) and (ii)

= $\frac{0.1}{4} \times \frac{R}{0.67} d T$

$=\frac{m}{2} \times \frac{R}{0.4} \times d T$

$m=\frac{0.1}{2} \times \frac{0.4}{0.67}$

$m=0.0298 \approx 0.03 g$

Therefore  the mass of hydrogen 0.03 g in one vessel and 0.1 g of helium in another vessel with adiabatic walls, when they are electrically heated  with equal amount of heat, it is found that temperature rise also the same .