Question

When the temperature of an ideal gas is increased from $27?C$ to $927?C$, the kinetic energy will be

• A. same
• B. eight times
• C. four times
• D. twice.

Answer 1

Answer: (C)

four times

Answer 2

Temperature of ideal gas is increased from $27?C$ to $927?C$
$\therefore \: KE =\frac{3}{2} RT$
$T_{1}=27+273 \Rightarrow 300 K$
$T_{2}=927 + 273 \Rightarrow 1200 K$
$\therefore \:\:\: KE_{1}=\frac{3}{2} \times R \times300 K$
$KE_{2} = \frac{3}{2}\times R \times1200 K$
So, $\frac{KE_{2}}{KE_{1}} =\frac{\frac{3}{2}\times R \times1200 K}{\frac{3}{2}\times R\times300 K}=4$
Hence, kinetic energy is increased four times.