In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq. A good estimate for q is :

Question

In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq. A good estimate for q is : (A) 2 (B) 1 (C) (1/2) (D) (1/4)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dc03a1cfa3c7cdec898fdeac9aeabb49-.png" /> Time interval between two successive collisions at a face =(2 l/u) Number of collisions/s =(u/2 l) Change of momentum/collision =2 m u Therefore, change of momentum/s =(u/2 l) × 2 m u=(m u2/l)=F Therefore, F propto u2 propto T ⇒ q=1

Q. In an ideal gas at temperature $T$ , the average force that a molecule applies on the walls of a closed container depends on $T$ as $T^{q}$ . A good estimate for $q$ is :

$2$

$1$

$\frac{1}{2}$

$\frac{1}{4}$

Time interval between two successive collisions at a face $= \frac{2 l}{u}$
Number of collisions/s $= \frac{u}{2 l}$
Change of momentum/collision $= 2 m u$
Therefore, change of momentum/s $= \frac{u}{2 l} \times 2 m u = \frac{m u ^{2}}{l} = F$
Therefore, $F \propto u^{2} \propto T \Rightarrow q = 1$

Q. In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq. A good estimate for q is :

2

1

21​

41​

Time interval between two successive collisions at a face =u2l​ Number of collisions/s =2lu​ Change of momentum/collision =2mu Therefore, change of momentum/s =2lu​×2mu=lmu2​=F Therefore, F∝u2∝T⇒q=1

Q. In an ideal gas at temperature $T$ , the average force that a molecule applies on the walls of a closed container depends on $T$ as $T^{q}$ . A good estimate for $q$ is :

$2$

$1$

$\frac{1}{2}$

$\frac{1}{4}$