Question 722: Thermodynamics & KTG

Question

The temperature of a gas is $-80^{\circ} \mathrm{C}$. To what temperature the gas should be heated so that the r.m.s. speed is increased by 2 times?

Accepted Answer

To solve the problem, we begin by converting the initial temperature from Celsius to Kelvin:
$ T_1 = -80^{\circ} \mathrm{C} = -80 + 273 \, \mathrm{K} = 193 \, \mathrm{K} $
The root mean square (RMS) velocity ($v_{\mathrm{rms}}$) of a gas is proportional to the square root of its temperature, $T$:
$ v_{\mathrm{rms}} \propto \sqrt{T} $
When the speed is increased by two times, the final speed becomes:
$ (2 \, v_{\mathrm{rms}})_1 $
This implies:
$ \frac{(v_{\mathrm{rms}})_2}{(v_{\mathrm{rms}})_1} = \sqrt{\frac{T_2}{T_1}} = 2 $
Therefore, the relationship between the initial and final temperatures is:
$ \frac{T_2}{T_1} = 4 $
Solving for $T_2$:
$ T_2 = 4 	imes 193 \, \mathrm{K} = 772 \, \mathrm{K} $
Converting back to Celsius, we subtract 273:
$ T_2 = 772 \, \mathrm{K} - 273 = 499^{\circ} \mathrm{C} $
Thus, the gas should be heated to $499^{\circ} \mathrm{C}$ to double the RMS speed.

Answer

$499^{\circ} \mathrm{C}$

Answer

$772^{\circ} \mathrm{C}$

Answer

$1464^{\circ} \mathrm{C}$

Answer

$1737^{\circ} \mathrm{C}$

Thermodynamics & KTG

The temperature of a gas is -80^{\circ} \mathrm{C}. To what temperature the gas should be heated so that the r.m.s. speed is increased by 2 times?

Choose the correct answer: