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:
$U = 4 \times \frac{5}{2} \times 8.31 \times 300$
$U = 2 \times 5 \times 8.31 \times 300$
$U = 10 \times 2493 = 24930 \ J$
$U = 24.93 \ kJ$.