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4

It may be helpful to look at a related value $k_{B}$, the Boltzmann constant, which is widely used in thermodynamics. These two are related by $R = k_{B}N_{A}$, allowing the ideal gas law to also be written: $$PV = Nk_{B}T$$ where $N$ is the number of particles, as opposed to the number of moles. The units are $\pu{J\cdot K^{-1}}$. It's a proportionality ...

2

Charles' law says that at constant pressure the volume and temperature of an ideal gas are related as $$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$ If $V_2=V_1+dV$ and $T_2=T_1+dT$ then $$\frac{V_1}{T_1}=\frac{V_1+dV}{T_1+dT}=\frac{V_1}{T_1}\left(\frac{1+dV/V_1}{1+dT/T_1}\right)$$ which can be rearranged into $$\frac{dV}{dT}=\frac{V_1}{T_1}$$ But Charles' law says ...

4

This definition is pretty ambiguous, and in my opinion not very helpful. But if you consider an ideal gas, for which we have $$pV = nRT$$ then $nRT$ can be identified with something which has the dimensions of p-V work. The problem is that 'work' is a path function, which must be specified with respect to a process, which goes from an initial state to a ...

2

Some notes I had easily at hand may help. You have already calculated the probability of obtaining the chance with a number $k$ of type of ball (or molecule) out of a total on $n$ is $$\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n} \tag{25c}$$ This distribution is a maximum when $k=n/2$. This can be seen with a straightforward argument. The factorial terms ...

1

Assumptions: Reaction of carbon dioxide with water is neglected. Vapour pressure of water is negligible. Volume of solution does not change on dissolution of carbon dioxide. Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2_{(aq.)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$ Initially no $\ce{CO2}$ ...

2

$B=\pu{0.6226 bar}$ as the water-water vapour system would be in equilibrium. Now, $A+B= \pu{1 bar}\\ A= \pu{0.3774 bar}$ For initial volume of the container, $P_{\ce{Ar}_i}V_i=n_{\ce{Ar}_i}RT\\ V_i=C=\frac{n_{\ce{Ar}_i}RT}{P_{\ce{Ar}_i}}\\ C=\pu{\frac{0.1 \times 0.08314 \times 360}{0.3774} L}\\ C=\pu{7.93 L}\\$ For initial moles of water vapour, $P_{\ce{... 3 OP has given a good effort to solve the problem using correct path. Only loose point was not considering the diatomic nature of the gases as Safdar Faisal pointed out in a comment. Suppose the amount of$\ce{O2}$in the gas mixture is$n_1 \ \pu{mol}$and the amount of$\ce{N2}$in the gas mixture is$n_2 \ \pu{mol}$in volume of$\pu{1.0 L}\$ container. Thus,...

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