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

Correct answer: C. four times Increase in temperature of ideal gas from 27^ C to 927^ C Increase in K E= (3)/(2) R T aligned T_1 & =27+273 & =300 K T_2 & =927+273 =1200 K K E_1 & =(3)/(2) × R × 300 K K E_2 & =(3)/(2) × R × 1200 K So, (K E_2)/(K E_1) & = (3)/(2) × R × 1200 K (3)/(3) × R × 300 K =4 aligned Hence kinetic energy is increased four times.

JIPMER 2005

Exam: JIPMER
Year: 2005
Subject: Chemistry
Chapter: States of Matter

JIPMER 2005 · Chemistry · States of Matter

When the temperature of an ideal gas is increased from 27^ C to...

When the temperature of an ideal gas is increased from

27^{\circ} \mathrm{C}

927^{\circ} \mathrm{C}

, the kinetic energy will be

Options

Asame
Beight times
Cfour times
Dtwice

Correct answer

C. four times

Step-by-step solution

Increase in temperature of ideal gas from

27^{\circ} \mathrm{C}

to 927^{\circ}

\mathrm{C} \therefore

Increase in

K E= \frac{3}{2} R T

\begin{aligned} T_1 & =27+273 \\ & =300 \mathrm{ K} \\ T_2 & =927+273 \Rightarrow=1200 \mathrm{ K} \\ \therefore \quad K E_1 & =\frac{3}{2} \times R \times 300 \mathrm{ K} \\ K E_2 & =\frac{3}{2} \times R \times 1200 \mathrm{ K} \\ \text{ So, } \frac{K E_2}{K E_1} & =\frac{\frac{3}{2} \times R \times 1200 \mathrm{ K}}{\frac{3}{3} \times R \times 300 \mathrm{ K}}=4\end{aligned}

$$ Hence kinetic energy is increased four times.

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