Question

If the heat required to increase the rms speed of 4 moles of a diatomic gas from $v$ to $\sqrt{3}v$ is $83.1\,\mathrm{kJ}$, then the initial temperature of the gas is (Universal gas constant $= 8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$)

• A. $377\,^\circ\mathrm{C}$
• B. $327\,^\circ\mathrm{C}$
• C. $227\,^\circ\mathrm{C}$
• D. $277\,^\circ\mathrm{C}$

Hint:

RMS speed depends only on temperature ($v_{rms} = \sqrt{3RT/M}$). Increasing speed by factor $k$ requires temperature to increase by factor $k^2$.

Answer 1

The Correct Option is C

Step 1: Relate RMS Speed and Temperature:
$v_{rms} \propto \sqrt{T}$. Given $v_{initial} = v$ and $v_{final} = \sqrt{3}v$. \[ \frac{v_{final}}{v_{initial}} = \frac{\sqrt{3}v}{v} = \sqrt{3} \] \[ \sqrt{\frac{T_f}{T_i}} = \sqrt{3} \implies \frac{T_f}{T_i} = 3 \implies T_f = 3T_i \] Change in temperature $\Delta T = T_f - T_i = 3T_i - T_i = 2T_i$.
Step 2: Heat Energy Calculation:
For "heat required to increase rms speed", we consider the change in internal energy ($\Delta U$) or heat at constant volume ($Q_v$), as speed is a function of internal kinetic energy. For a diatomic gas, $C_v = \frac{5}{2}R$. \[ Q = n C_v \Delta T \] Given $n = 4$ moles, $Q = 83.1\,\mathrm{kJ} = 83100\,\mathrm{J}$. \[ 83100 = 4 \times \frac{5}{2} R \times (2T_i) \] \[ 83100 = 4 \times \frac{5}{2} \times R \times 2T_i = 20 R T_i \]
Step 3: Solve for Initial Temperature:
\[ 20 \times 8.31 \times T_i = 83100 \] \[ 166.2 T_i = 83100 \] \[ T_i = \frac{83100}{166.2} = 500\,K \]
Step 4: Convert to Celsius:
\[ T_i = 500 - 273 = 227\,^\circ\mathrm{C} \]