The problem states that the piston can slide without friction and does not move after closing the switch. This implies that the system is in mechanical equilibrium at all times. Therefore, the pressure of Hydrogen gas ($P_1$) must equal the pressure of Helium gas ($P_2$): $$ P_1 = P_2 $$ Using the Ideal Gas Law $PV = nRT$, we can write: $$ \frac{n_1 R T_1}{V_1} = \frac{n_2 R T_2}{V_2} $$ Rearranging this to find the relationship between the temperatures of the two gases: $$ T_1 = \left( \frac{n_2 V_1}{n_1 V_2} \right) T_2 \quad \dots \text{(i)} $$

The heating coils are connected in series, so the same current $I$ flows through both. The heat generated in a resistor is $dQ = I^2 R dt$. Since the piston is stationary ($dV = 0$), the process is isochoric for both gases. The heat supplied raises the internal energy: $$ dQ = dU = n C_v dT $$ For Hydrogen (Gas 1): $$ n_1 C_{v1} dT_1 = I^2 R_1 dt \quad \dots \text{(ii)} $$ For Helium (Gas 2): $$ n_2 C_{v2} dT_2 = I^2 R_2 dt \quad \dots \text{(iii)} $$

Dividing equation (ii) by equation (iii): $$ \frac{n_1 C_{v1} dT_1}{n_2 C_{v2} dT_2} = \frac{R_1}{R_2} $$ From equation (i), we know that $T_1$ is proportional to $T_2$. Differentiating equation (i) gives: $$ dT_1 = \left( \frac{n_2 V_1}{n_1 V_2} \right) dT_2 $$ Substituting this into the ratio: $$ \frac{R_1}{R_2} = \frac{n_1 C_{v1}}{n_2 C_{v2}} \cdot \left( \frac{n_2 V_1}{n_1 V_2} \right) $$ $$ \frac{R_1}{R_2} = \frac{C_{v1} V_1}{C_{v2} V_2} $$ Solving for $R_2$: $$ R_2 = R_1 \left( \frac{C_{v2} V_2}{C_{v1} V_1} \right) \quad \dots \text{(iv)} $$

Hydrogen ($H_2$) is a diatomic gas, so $C_{v1} = \frac{5}{2}R$.
Helium ($He$) is a monoatomic gas, so $C_{v2} = \frac{3}{2}R$.
Substituting these into the ratio: $$ \frac{C_{v2}}{C_{v1}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5} $$ Therefore, equation (iv) becomes: $$ R_2 = R_1 \left( \frac{3 V_2}{5 V_1} \right) $$

We are given $R_1 = R_0 + A T_1$. We need to find the law for $R_2$ as a function of the Helium temperature $T$ (where $T = T_2$). First, express $T_1$ in terms of $T$ using equation (i): $$ T_1 = T \frac{n_2 V_1}{n_1 V_2} $$ Using the molar mass relation $n = \frac{m}{M}$: $$ \frac{n_2}{n_1} = \frac{m_2 / M_{He}}{m_1 / M_{H_2}} = \frac{m_2 M_{H_2}}{m_1 M_{He}} $$ So, $$ T_1 = T \left( \frac{m_2 M_{H_2} V_1}{m_1 M_{He} V_2} \right) $$ Now substitute $T_1$ into the expression for $R_1$, and then $R_1$ into the expression for $R_2$: $$ R_2 = \left[ R_0 + A \left( T \frac{m_2 M_{H_2} V_1}{m_1 M_{He} V_2} \right) \right] \left( \frac{3 V_2}{5 V_1} \right) $$