The molecules of a given mass of a gas have r.m.s velocity of 200 ms-1 at 27° C and 1.0 × 105 Nm-2 pressure. When the temperature and pressure of the gas are respectively, 127° C and 0.05 × 105 Nm-2, the r.m.s. velocity of its molecules in ms-1 is

Question

The molecules of a given mass of a gas have r.m.s velocity of 200 ms-1 at 27° C and 1.0 × 105 Nm-2 pressure. When the temperature and pressure of the gas are respectively, 127° C and 0.05 × 105 Nm-2, the r.m.s. velocity of its molecules in ms-1 is (A) (400/√3) (B) (100 √2/3) (C) (100/3) (D) 100 √2

Tardigrade Admin · Accepted Answer

v propto √T ⇒ (v/200) = √(400/300) ⇒ v = (200 ×2/√3) m/s ⇒ v = (400/√3) m/s

$\frac{400}{3}$

$\frac{100 2}{3}$

$\frac{100}{3}$

$1002$

$v \propto T \Rightarrow \frac{v}{200} = \frac{400}{300} \Rightarrow v = \frac{200 \times 2}{3} m / s$
$\Rightarrow v = \frac{400}{3} m / s$

Q. The molecules of a given mass of a gas have r.m.s velocity of 200ms−1 at 27∘C and 1.0×105Nm−2 pressure. When the temperature and pressure of the gas are respectively, 127∘C and 0.05×105Nm−2, the r.m.s. velocity of its molecules in ms−1 is

3​400​

31002​​

3100​

1002​

v∝T​⇒200v​=300400​​⇒v=3​200×2​m/s ⇒v=3​400​m/s

$\frac{400}{3}$

$\frac{100 2}{3}$

$\frac{100}{3}$

$1002$

$v \propto T \Rightarrow \frac{v}{200} = \frac{400}{300} \Rightarrow v = \frac{200 \times 2}{3} m / s$
$\Rightarrow v = \frac{400}{3} m / s$