Question

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 387 K to 774 K. If v1 is the average speed of the gas molecules before the absorption of heat and v2 their average speed after the absorption of heat, what is the ratio v2 v1 ? 1. v2 v1 = √ 2 2. v2 v1 = 2 3. v2 v1 = 1 2 4. v2 v1 = 4 5. v2 v1 = 1

Answer 1

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Answer 2

Answer:

Ratio v₂/v₁ = √2

Explanation:

We are given that ;

Initial temperature (T1) = 387K

Final temperature (T2) = 774K

Now, according to kinetic theory of gases, the average speed is proportional to the square root of the absolute temperature. The speed is given by:

v = √( 3∙R∙T/M)

Where;

R = universal gas constant,

T = absolute temperature,

M = molar mass of the gas molecules

Thus, v₁ = √(3∙R∙T₁/M)

Since it's same gas,

v₂ = √( 3∙R∙T₂/M)

Hence,

Ratio v₂/v₁ = √( 3∙R∙T₂/M)/√( 3∙R∙T₁/M) = √(T₂/T₁) = √(774K/387K) = √2