When the temperature of a gas is raised from 27° C to 90° C, the percentage increase in the rms velocity of the molecules will be

Question

When the temperature of a gas is raised from 27° C to 90° C, the percentage increase in the rms velocity of the molecules will be (A) 10 % (B) 15 % (C) 20 % (D) 17.5 %

Tardigrade Admin · Accepted Answer

(C'/C)=√(273+90/273+27)=√(363/300) or (C'/C)=√1.21=1.1 or (C'/C)-1=0.1 or (C'-C/C) × 100=0.1 × 100=10

Q. When the temperature of a gas is raised from $2 7^{\circ} C$ to $9 0^{\circ} C$ , the percentage increase in the rms velocity of the molecules will be

$10%$

$15%$

$20%$

$17.5%$

$\frac{C ^{'}}{C} = \frac{273 + 90}{273 + 27} = \frac{363}{300}$
or $\frac{C ^{'}}{C} = 1.21 = 1.1$
or $\frac{C ^{'}}{C} - 1 = 0.1$
or $\frac{C ^{'} - C}{C} \times 100 = 0.1 \times 100 = 10$

Q. When the temperature of a gas is raised from 27∘C to 90∘C, the percentage increase in the rms velocity of the molecules will be

10%

15%

20%

17.5%