The kinetic energy of an ideal gas at 27^{cir | vedclass

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Question

(A) The average kinetic energy of an ideal gas is given by $E = \frac{3}{2}kT$,which implies $E \propto T$.First,convert the temperatures from Celsius

Vedclass · Accepted Answer

$2E_1$

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(A) The average kinetic energy of an ideal gas is given by $E = \frac{3}{2}kT$,which implies $E \propto T$.First,convert the temperatures from Celsius to Kelvin:$T_1 = 27 + 273 = 300 \ K$$T_2 = 327 + 273 = 600 \ K$Using the proportionality $E \propto T$,we have:$\frac{E_2}{E_1} = \frac{T_2}{T_1}$$\frac{E_2}{E_1} = \frac{600 \ K}{300 \ K} = 2$Therefore,$E_2 = 2E_1$.

The kinetic energy of an ideal gas at $27^{\circ}C$ is $E_1$. If the temperature is increased to $327^{\circ}C$,what will be the new kinetic energy?

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