Pressure and Energy Questions in English

(A) For an ideal gas,the intermolecular forces are assumed to be negligible,and collisions are perfectly elastic.
Consequently,the internal energy $U$ of an ideal gas is solely a function of its absolute temperature $T$.
According to the kinetic theory of gases,the internal energy of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature in Kelvin.
Since $U \propto T$,the relationship between internal energy $U$ and temperature $T$ is linear,passing through the origin $(0, 0)$.
Therefore,the graph that best illustrates this relationship is a straight line passing through the origin,which corresponds to Graph $A$.

(C) The average translational kinetic energy $(K_{avg})$ of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
Since the temperature $T$ is the same for both hydrogen and oxygen, the average translational kinetic energy depends only on the temperature.
Therefore, the ratio of the average translational kinetic energies of hydrogen and oxygen is $1: 1$.

(A) The pressure of an ideal gas is given by the kinetic theory formula:
$p = \frac{1}{3} \frac{M}{V} v_{rms}^2 = \frac{1}{3} \frac{N m}{V} v^2$
where $m$ is the mass of a molecule, $N$ is the number of molecules, $V$ is the volume, and $v$ is the root-mean-square speed.
Let the initial pressure be $p = \frac{1}{3} \frac{N m}{V} v^2$.
When the mass of each molecule is doubled $(m' = 2m)$ and the speed is halved $(v' = v/2)$, the new pressure $p'$ is:
$p' = \frac{1}{3} \frac{N (2m)}{V} \left(\frac{v}{2}\right)^2$
$p' = \frac{1}{3} \frac{N (2m)}{V} \left(\frac{v^2}{4}\right) = \frac{1}{2} \left( \frac{1}{3} \frac{N m}{V} v^2 \right) = \frac{1}{2} p$
Therefore, the ratio of initial pressure to final pressure is:
$\frac{p}{p'} = \frac{p}{p/2} = \frac{2}{1}$
Thus, the ratio is $2: 1$.

(C) The formula for the total internal energy $U$ of an ideal gas is given by $U = n \frac{f}{2} R T$.
For a diatomic gas, the degrees of freedom $f = 5$.
Given: number of moles $n = 4$, temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$, and $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
Substituting these values into the formula:
$U = 4 \times \frac{5}{2} \times 8.31 \times 300$
$U = 2 \times 5 \times 8.31 \times 300$
$U = 10 \times 2493 = 24930 \ J$
$U = 24.93 \ kJ$.

(A) For an ideal gas, the internal energy $U$ is related to pressure $P$ and volume $V$ by the formula: $U = \frac{3}{2} PV$.
Given, energy per unit volume (energy density) is $u = \frac{U}{V} = 600 \text{ J/L}$.
Since $1 \text{ L} = 10^{-3} \text{ m}^3$, the energy density in $SI$ units is $u = \frac{600 \text{ J}}{10^{-3} \text{ m}^3} = 6 \times 10^5 \text{ J/m}^3$.
Using the relation $u = \frac{3}{2} P$, we get $P = \frac{2}{3} u$.
Substituting the value of $u$: $P = \frac{2}{3} \times 6 \times 10^5 \text{ N/m}^2 = 4 \times 10^5 \text{ N/m}^2$.

(C) According to the kinetic theory of gases, the pressure $(P)$ exerted by an ideal gas on the walls of a container is given by the relation $P = \frac{1}{3} \rho v_{rms}^2$, where $\rho$ is the density and $v_{rms}$ is the root mean square velocity.
Since $v_{rms} = \sqrt{\frac{3RT}{M}}$, we have $v_{rms}^2 = \frac{3RT}{M}$.
Substituting this into the pressure equation, we get $P = \frac{1}{3} \rho \left(\frac{3RT}{M}\right) = \frac{\rho RT}{M}$.
Since the density $\rho$ and molar mass $M$ are constant for a fixed amount of gas in a closed container, the pressure $P$ is directly proportional to the temperature $T$ $(P \propto T)$.
Force $(F)$ is defined as pressure multiplied by the area of the wall $(F = P \times A)$.
Since the area of the container walls is constant, the force $F$ is also directly proportional to the pressure, and thus directly proportional to the temperature $(F \propto T^1)$.

(D) The translational kinetic energy $(K)$ of a gas molecule is given by the formula: $K = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
From this relation, it is clear that $K \propto T$.
If the kinetic energy is to be doubled, i.e.,$K' = 2K$, then the new temperature $T'$ must satisfy the relation $K' \propto T'$.
Since $K' = 2K$, we have $T' = 2T$.
Therefore, the absolute temperature must be increased to $2T$ to double the translational kinetic energy of the gas molecules.

(C) The kinetic energy $(K.E.)$ of an ideal gas is directly proportional to its absolute temperature $(T)$ in Kelvin.
$K.E. \propto T$
Given, initial temperature $T_1 = 27^\circ$$C$ = $27 + 273 = 300$ $K$.
Initial kinetic energy $K_1 = E_0$.
Final temperature $T_2 = 177^\circ$$C$ = $177 + 273 = 450$ $K$.
Using the relation $\frac{K_2}{K_1} = \frac{T_2}{T_1}$:
$\frac{K_2}{E_0} = \frac{450}{300}$
$\frac{K_2}{E_0} = \frac{3}{2}$
$K_2 = \frac{3}{2} E_0$.

(A) According to the kinetic theory of gases, the pressure $P$ exerted by an ideal gas is given by the expression:
$P = \frac{1}{3} \rho v_{rms}^2$
where $\rho$ is the density of the gas and $v_{rms}$ is the root mean square velocity.
We know that the kinetic energy per unit volume $E$ is given by:
$E = \frac{1}{2} \rho v_{rms}^2$
From this, we can write $\rho v_{rms}^2 = 2E$.
Substituting this into the pressure expression:
$P = \frac{1}{3} (2E) = \frac{2}{3} E$
Thus, the pressure is $(2/3)$ rd of the kinetic energy per unit volume of the gas.