13

It's not a coincidence at all! If you do an online search for "derivation of osmotic pressure", you'll see how $R$ enters into the derivation. Indeed, that's one of the beauties of the van 't Hoff equation for osmotic pressure – it reveals that (under the modest simplifying assumptions of the van 't Hoff equation derivation) the osmotic pressure created ...


11

The answer to your question is yes and no. You are correct in supposing that as an ideal gas expands the entropy will increase as it has more space to occupy and so the number of ways the molecules can be placed in the total space has increased. (Try to not think of entropy in terms of randomness but in the number of ways molecules can be positioned in space,...


10

Here's your confusion: You need to consider two different things: The momentum transfer per particle per collision. There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that ...


5

In ideal gases no intermolecular forces, therefore no potential energy. Thus, internal energy is equal to total kinetic energy (KE) of the system. Consider $N$ monoatomic particles in a cubical box of side $\ell$ (assumption: ideal gasses consist of monoatomic point particles). The amount of ideal gas in the box is $\frac{N}{N_A} = n \ \pu{mol}$ where $N_A$ ...


4

His name was Amedeo Avogadro, not Amidio nor Amadeo nor Avegadro nor Avagadro. The full name was Lorenzo Romano Amedeo Carlo Avogadro, Count of Quaregna and Cerreto. Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. More exactly, it also includes single atoms, e.g. for a case of ...


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 ...


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 ...


3

Because it gives simpler-to-derive laws which are often very good approximations Clearly real gases do not always follow the ideal gas laws. They mostly liquefy under some conditions, for example, and, under those conditions they are clearly not ideal. But in practice gas laws are used for things far away from those non-ideal regions. When we are applying ...


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,...


3

In general for a perfect or ideal gas, $$C_p=C_V + R'$$ (using your notation) where the heat capacities are molar quantities. It follows that for a perfect gas mixture $(C_p)_\text{mix}=(C_V)_\text{mix} + R'$.


3

When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you ...


2

Internal energy of any gas is given by $$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$ Last three aren't affected by ordinary heating. And $U_\text{intermolecular}$ for ideal gas is zero. That's why for ideal gas $U$ is only function of temperature. But ...


2

The temperature and the volume of the inner ear are constant. When your ears pop during descent, air from the cabin goes into the ear, increasing the pressure. The law is the following: $$n / P = const$$ You can derive this from the ideal gas law. It has no special name.


2

To carry out the B transformation, you have to heat a lot. If you don't heat, the volume will of course increase, but the pressure will decrease. B is a process difficult to carry out, because it is not easy to heat enough a gas who is inflated to as to maintain the inner pressure. Then, at the end of B, the temperature is very high. You block the position ...


2

Your analysis is correct in terms of number density. But let's see how it plays out in terms of molar density. Let n be the number of moles and A be Avagadro's number. Then N=nA. If we substitute this into your first equation, then I get $$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where ...


2

We know from the first Newton motion law, that the net force acting on an object in rest must be zero. The forces acting on the piston are gravity and gas pressure: $$\vec F_g + \vec F_\mathrm{p,down} + \vec F_\mathrm{p,up}=\vec 0 \tag{1}$$ If $V$ is the given bottom gas volume, $V_0$ is the total gas volume, $n$ is the molar amount of each of gases, the gas ...


2

The Gibbs free energy change is zero in the case of reversible processes carried out at constant temperature and pressure, but that isn't the case if these conditions are not observed. As a demonstration consider an isothermal reversible expansion of an ideal gas. Since the temperature is constant, the free energy change is given by $$\Delta G = \Delta H - T ...


2

T in K $V^2$ in $\pu{m2 s-2}$ 300 206116 323 230400 363 260100 403 280900 498 348100 623 435600 698 504100 773 547600 These are proportional.


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{...


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 ...


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 ...


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 $\ce{CO2}$ ...


1

Your claim that $\Delta H = \Delta U + P_{ext} \Delta V$ is incorrect. It is only correct when the system is always in mechanical equilibrium with the external pressure, or in other words irreversible expansion against constant pressure. The more general statement is $\Delta H = \Delta U + \Delta (PV) $. Here as you can see even when you compare it with $\...


1

Cp of an ideal gas doesn't depend on pressure. Cp of any substance is defined as the partial derivative of enthalpy respect to temperature at constant pressure. But, if the enthalpy of the substance is independent of pressure, then it doesn't matter if the pressure is constant. However, determining Cp by measuring the amount of heat required to change the ...


1

At the interface between the gas and its surroundings, by Newton's 3rd law, the pressure exerted by the gas on its surroundings is equal to the pressure exerted by the surroundings on the gas. So we can use either. For a reversible expansion, the gas pressure is given by the ideal gas law. But the ideal gas law only applies at thermodynamic equilibrium ...


1

When real gases are at high temperature, the kinetic energy prevents any gas particles from interacting via intermolecular forces. With low pressure, the gas particles are separated enough that the intermolecular forces are sparse, therefore, giving rise to the ideal behavior since ideal gases are defined as non-interacting particles. When real gases are at ...


1

The van Der Waal equation is $$\left(p+\frac a{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$ Here $V_\mathrm m$ is molar volume. When pressure is low and temperature is very high, we can qualitatively say that the molar volume will be very large. Due to this the volume occupied by the molecules (given by $b$) becomes insignificant. The pressure is low but the ...


1

There are at least two fundamental issues you have to address. First, you have to distinguish between the Gibbs energy $G$ and the Gibbs energy of reaction $\Delta_r G$. In you diagram, one is the value on the y-axis (without defined zero point) and the other is the slope of the line, as labeled in your sketch. The second issue is that in the expressions on ...


1

I can't see what you did in your derivation, but, for what it's worth, here's my derivation. My starting equations are $$G=\sum{n_i\mu_i}$$and, along your contour, at constant temperature and pressure, $$\mathrm dG=\sum{\mu_i\mathrm dn_i}$$The changes in the number of moles of the various species are given by $$n_i=n_{i0}+c_in$$where $c_i$ is the ...


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