Tag: behaviour of perfect gas and kinetic theory of gases

At a depth of h water pressure will be
${P} _{h} = {P} _{a} + \rho gh$                eq(1)
where 
$\rho  =1000 kg/{m}^{3}$
$g  = 9.81 m/{s}^{2} $
$h = \text{depth  of  water  bubble}$
${P} _{a} = {10}^{5}Pa$

initially bubble is below water at h, so
${P} _{1} = {P} _{h}$
finally it rises to surface of lake so
${P} _{2} = {P} _{a}$

we assume that temperature is constant when it is rising from bottom to surface
by ideal gas equation
${P} _{1}{V} _{1} = {P} _{2}{V} _{2}$

volume can be written as $V = \dfrac{4}{3}\pi {r}^{3}$
where r is radius.
${P} _{h} \times \dfrac{4}{3}\pi {r} _{h}^{3} = {P} _{a} \times \dfrac{4}{3}\pi {r} _{s}^{3}$
${r} _{h} = {(\dfrac{{P} _{a}}{{P} _{h}})}^{1/3} r$
${r} _{h} = \dfrac{r}{{(\frac{{P} _{h}}{{P} _{a}})}^{1/3}}$
by eq(1)
${r} _{h} = \dfrac{r}{{(\dfrac{{P} _{a} +  \rho gh}{{P} _{a}})}^{1/3}}$
${r} _{h} = \dfrac{r}{{(1 + 0.0981h)}^{1/3}}$              eq(2)

(a) at depth 30m
h = 30m
put h=30 in eq(2)
${r} _{h} = \dfrac{r}{1.5}$
false

(b) at depth 70m
h = 70m
put h=70 in eq(2)
${r} _{h} = \dfrac{r}{2}$
true

(c) at depth 140m
h = 140m
put h=140 in eq(2)
${r} _{h} = \dfrac{r}{2.4}$
false

(d) at depth 260m
h = 260m
put h=260 in eq(2)
${r} _{h} = \frac{r}{3}$
true

Answer is D.

Let the new pressure inside be $P$ after air introduced.

(a) Temperature must be constant to apply boyle's law
(b)From vanderwaal's equation when non idealities of a gas is undertaken then we can apply boyle's only when temperture is high and pressure is low
(c)Vessel enclosing must be a good conductor so there is no any possibilities of adiabatic process
(d) Temperature must be constant therefore, process must be isothermal
Hence all are correct
Hence option(D)

Using v = sqrt(gamma * P / rho) and P = (rho/M)RT, we find gamma. Since v = 330 m/s, rho = 1.3 kg/m^3, and P = 1.013e5 Pa, gamma = v^2 * rho / P = 1.71. However, for standard gases, gamma = 1 + 2/f. With gamma = 1.4 (diatomic), f = 5. The provided data yields gamma approx 1.4, corresponding to f = 5.

According to the Stefan-Boltzmann law, radiated energy E is proportional to T^4. If T increases by 10%, T_new = 1.1T. Then E_new = (1.1)^4 * E = 1.4641 * E. The increase is 46.41%.

The law is given by Claussius which states that for any dynamical system in a thermal equilibrium, the total energy is equally divided among the degrees of freedom.

Given that value of gamma is 1.66 i.e $\dfrac{5}{3}$ which implies that it is a monoatomic gas, and Neon (Ne) is the only monoatomic gas among the given options.

Degrees of freedom represent the number of independent coordinates required to specify the position of an object. An ant on a 2D plane requires two coordinates (x, y).

Gas molecules are in random motion having some momentum and while colliding with the walls they transfer their momentum to the walls and this collective transfer of momentum from all the molecules to the walls appears as pressure exerted by gas on the container wall.