Explanation
$$T=-\alpha V^3+\beta V^2\\ \cfrac{PV}{R}=-\alpha V^2+\beta V^2\\ P=RV(-\alpha V+\beta)$$
For maximum pressure
$$\cfrac{dP}{dV}=0\\ \Rightarrow -2\alpha V+\beta =0\\ \Rightarrow V=\cfrac{\beta }{2\alpha}\\ P=R[-\alpha(\cfrac{\beta }{2\alpha})^2+\cfrac{\beta }{2\alpha}\times \beta]\\P=R[\cfrac{-\beta^2}{4\alpha}+\cfrac{\beta^2}{2\alpha}]\\P=\cfrac{R\beta^2}{4\alpha}$$
Ideal gas law, says that
$$PV = N K T$$.
This means that:
Average Kinetic Energy, $$K_{avg} = \dfrac{3}{2} k T$$
This is a very important fact, as it tells the fundamentals about the temperature. The absolute temperature of the ideal gas is directly proportional to the average kinetic energy. So, changes in pressure and/or volume will result in changes in temperature too. Also, we can see that the average kinetic energy of the ideal gas
molecule will depend on temperature only and not on the gas nature/mass. Therefore, if the temperature is the same, then the kinetic energy of $$O_{2} $$ and $$N_{2}$$ both will be equal.
Hence average K.E. for $$N_{2}$$ will also be 0.768 eV.
Option D is correct.
Let V is the volume of the vessel.
Now, let $$p_{1}$$ and $$p_{2}$$ be the partial pressure, then using gas law:
$$p_{1}V = \dfrac{m_1}{M_1}RT\\$$
$$p_{2}V = \dfrac{m_2}{M_2}RT,\ p_{0} = p_{1} + p_{2}\\$$
$$p_{0} = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{V}\\$$
$$V = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{p_{0}}\\$$
$$\because \rho_{mix}=\dfrac{(m_{1} + m_{2})}{V}\\$$
$$\rho_{mix}=\dfrac {(m_1 + m_2)M_1 M_2} {(m_1M_2 + m_2M_1)} \times \dfrac{p_0}{RT}\\$$
Substituting values,
$$\rho_{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$$
$$= 1.446 \ per \ litre$$
Option A is correct.
$$\textbf{Correct option: Option (A).}$$
$$\textbf{Explanation:}$$
Given, the pressure at melting point is 2.20 times the pressure at triple point of water.
$${{P}_{MP}}=2.20\times {{P}_{TP}}$$
We know that, Melting point, $$MP=\dfrac{{{P}_{MP}}}{{{P}_{TP}}}\times 273.16K$$
$$\Rightarrow MP=\dfrac{2.20\times {{P}_{TP}}}{{{P}_{TP}}}\times 273.16\approx 600K$$
There is no information about gas so the translational kinetic energy is given as,
$$K = \dfrac{3}{2}RT$$
$$ = \dfrac{3}{2} \times 8.31 \times \left( {273 + 0} \right)$$
$$ = 3.4 \times {10^3}\;{\rm{J}}$$
Thus, the kinetic energy of a gas molecule is $$3.4 \times {10^3}\;{\rm{J}}$$.
In an experiment, 1.35 mol of oxygen (O2) are heated at constant pressure starting at 11.0ºC. How much heat must be added to the gas to double its volume?
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