The temperature of a gas is $-80^{\circ} \mathrm{C}$. To what temperature the gas should be heated so that…
- $499^{\circ} \mathrm{C}$
- $772^{\circ} \mathrm{C}$
- $1464^{\circ} \mathrm{C}$
- $1737^{\circ} \mathrm{C}$
Solution
As the speed is increased by two times, final speed becomes $\left(2 \mathrm{~V}_{\mathrm{rms}}\right)_1$ $\begin{array}{ll} & \frac{\left(v_{\mathrm{mss}}\right)_2}{\left(v_{\mathrm{ms}}\right)_1}=\sqrt{\frac{T_2}{T_1}}=2 \\ \therefore \quad & \frac{T_2}{T_1}=4 \\ \therefore \quad & T_2=4 \times 193=772 \mathrm{~K}=499^{\circ} \mathrm{C} \end{array}$
Asked in: MHT CET 2024 (09 May Shift 1)