Explanation:
If t is the time taken for the formation of ammonium chloride by the combination of NH3 and HCl, then
${r_{HCl}} = {{60} \over t}$ ; ${r_{N{H_3}}} = {{40} \over t}$
The rate of diffusion, r is proportional to the root mean square velocity (u) which is equal to $\sqrt {{{3P} \over d}} $.
$r \propto u \propto {{{3P} \over d}} $
$\therefore$ ${{{r_{HCl}}} \over {{r_{N{H_3}}}}} = {{60/t} \over {40/t}} = {{\sqrt {{{3{P_{HCl}}} \over {{d_{HCl}}}}} } \over {\sqrt {{{3{P_{N{H_3}}}} \over {{d_{N{H_3}}}}}} }}$
or ${{60} \over {40}} = {{\sqrt {{{3{P_{HCl}}} \over {18.25}}} } \over {\sqrt {{{3 \times 1\,atm} \over {8.5}}} }}$
or ${\left( {{{60} \over {40}}} \right)^2} = {{3{P_{HCl}}} \over {18.25}} \times {{8.5} \over {3\,atm}}$
or ${P_{HCl}} = {{60 \times 60} \over {40 \times 40}} \times {{18.25} \over {8.5}}$ atm = 4.83 atm
Explanation:
To find the total kinetic energy ratio for oxygen and hydrogen at a given temperature, we first note that kinetic energy (average) of a molecule can be represented using the formula for translational kinetic energy of an ideal gas, which is given by:
$ KE_{avg} = \frac{3}{2} k_B T $
where $ k_B $ is the Boltzmann constant and $ T $ is the temperature in Kelvin.
For a given number of molecules $ N $, the total kinetic energy $ KE_{total} $ can be expressed as:
$ KE_{total} = N KE_{avg} $
However, in order to find $ N $, which is the number of molecules, we need the number of moles, since $ N = nN_A $ where $ n $ is the number of moles and $ N_A $ is Avogadro’s number. We calculate $ n $ by dividing the mass of the gas $ m $ by its molar mass $ M $. This gives:
$ n = \frac{m}{M} $
For hydrogen (H2), the molar mass $ M_{H_2} = 2 $ g/mol and for oxygen (O2), the molar mass $ M_{O_2} = 32 $ g/mol. Given that both gases have a mass of 8 grams each, we can calculate the moles for each:
$ n_{H_2} = \frac{8 \text{ g}}{2 \text{ g/mol}} = 4 \text{ moles} $
$ n_{O_2} = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} $
The number of molecules $ N $ for each gas becomes:
$ N_{H_2} = n_{H_2} N_A = 4 N_A $
$ N_{O_2} = n_{O_2} N_A = 0.25 N_A $
Even though the mass of each gas is the same, the number of moles (and therefore the number of molecules) is different. However, the total kinetic energy for any ideal gas sample is still dependant on the temperature and the total number of molecules. Therefore, for each gas, substituting from the kinetic energy formula, we get:
$ KE_{total, H_2} = N_{H_2} KE_{avg} = 4 N_A \times \frac{3}{2}k_B T $
$ KE_{total, O_2} = N_{O_2} KE_{avg} = 0.25 N_A \times \frac{3}{2}k_B T $
The ratio of total kinetic energies $(H_2/O_2)$ thus becomes:
$ \frac{KE_{total, H_2}}{KE_{total, O_2}} = \frac{4 N_A \times \frac{3}{2}k_B T}{0.25 N_A \times \frac{3}{2}k_B T} = \frac{4 N_A}{0.25 N_A} = 16 $
Even though both samples are at the same temperature and have the same mass, hydrogen has more molecules contributing to its kinetic energy due to its significantly lower molecular weight compared to oxygen. Hence, under these conditions, the ratio of the total kinetic energy of hydrogen to oxygen is $ 16:1 $.
Explanation:
The rate of diffusion of a gas is inversely proportional to both the pressure of the gas and the square root of its molecular mass. This relationship is quantitatively expressed in Graham’s Law of Effusion. According to this law, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass, mathematically expressed as:
$ R \propto \frac{1}{\sqrt{M}} $Where R is the rate of diffusion or effusion and M is the molecular mass of the gas.
The rate of diffusion is also influenced by the pressure of the gas. At higher pressures, the rate of diffusion decreases. However, the relationship between gas pressure and the rate of diffusion is not as straightforward as the relationship with molecular mass, and it isn't as commonly discussed in simple terms as is Graham’s Law. Typically, the effect of pressure on the rate of diffusion is more significant in specific contexts, such as under varying atmospheric conditions.
In fundamental terms, a higher pressure implies more gas molecules are present, leading to increased collision frequency among them, which can slow down the overall rate at which individual molecules spread out or move through a medium.
Explanation:
The difference between the molar heat capacities at constant pressure ($C_p$) and constant volume ($C_v$) for an ideal gas is a constant value given by the gas constant $R$. This relationship is expressed mathematically as:
$ C_p - C_v = R $Here, $R$ is the universal gas constant, with a value of approximately 8.314 J/(mol·K).
The reason behind this relationship lies in the energy needed to perform work against the external pressure when the gas is heated at constant pressure, which does not happen in the case of heating at constant volume. When an ideal gas is heated at constant pressure, not only does the internal energy of the gas increase (which is also the case at constant volume) but some of the energy is also used to do work on the surroundings by expanding, which contributes to the larger value of $C_p$ compared to $C_v$. This difference is equal to the amount of work done per unit increase in temperature, which is quantitatively equivalent to the gas constant $R$.
Therefore, the expression indicates that the difference in heat capacity (which essentially is the difference in the ability to store thermal energy under different conditions) is fundamentally linked to the work done by the gas during expansion at constant pressure. The equation $C_p - C_v = R$ holds true for an ideal gas because it assumes no interactions between the molecules and that all processes are reversible.