TG EAMCET 2 May 2025 Shift 2 Paper

Step 1: How rms speed relates to temperatureThe root mean square (rms) speed of a gas, $V_{\text{rms}}$, increases as the square root of its temperature ( $T )$. So, $V_{\text{rms}} \propto \sqrt{T}$.Step 2: Setting up the ratioSince $V_{\text{rms}}^2 \propto T$, we can write:$\left[ \frac{(V_{\text{rms}})_1}{(V_{\text{rms}})_2} \right]^2 = \frac{T_1}{T_2}$Step 3: Substituting given valuesThe rms speed increases from $v$ to $\sqrt{3} v$. So:$\left( \frac{v}{\sqrt{3} v} \right)^2 = \frac{T_1}{T_2}$This simplifies to $\frac{1}{3} = \frac{T_1}{T_2}$, so $T_2 = 3 T_1$.Step 4: Calculating heat requiredThe heat $(Q)$ needed to raise the temperature is:$Q = n C_V \Delta T$For 4 moles of a diatomic gas, $C_V = \frac{5}{2} R$, so:$Q = 4 \times \frac{5}{2} R (T_2 - T_1)$We know $T_2 = 3 T_1$, so $\Delta T = 3 T_1 - T_1 = 2 T_1$.Step 5: Plugging in the numbersWe are given $Q = 83.1\,\text{kJ} = 83,100\,\text{J}$ and $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$.$83,100 = 4 \times \frac{5}{2} \times 8.31 \times 2 T_1$$4 \times \frac{5}{2} \times 2 = 20$, so:$83,100 = 20 \times 8.31 \times T_1$$20 \times 8.31 = 166.2\,\text{, so: }\,83,100 = 166.2 \times T_1$$T_1 = \frac{83,100}{166.2} = 500\,\text{K}$Step 6: Converting to Celsius$T_1\,\text{ in Celsius } = 500 - 273 = 227^{\circ}\,\text{C}$