Ideal gas Definition and 823 Threads

Solution from the textbook: My work: I constantly get 1.55kg. I also tried dividing the first and the second equation (pxV=m/M x R x T with different values). How did they come up with the equation in the solution? Also, I am sorry if I posted it in the wrong place and didn't follow the rules...

A cylinder contains an initial volume V1 = 1m^^3 of a perfect gas at initial pressure p1 = 1 bar, confined by a piston that is held in place by a spring. The gas is heated until its volume is doubled and the final pressure is 5 bar. Assuming that the mass of the piston is negligible and that the...

Hi, I didn't understand the maths involved in the below article in regard to temperature and ideal gas thermometer. If any member knows it, may reply me. If triple point of water is fixed at 273.16 K, and experiments show that freezing point of air-saturated water is 273.15 K at 1 atm...

Hi, I tried to do this question in two different approaches one of them was using the equation PV=mRT where I got the right answer which is 4.305 m**2. However, I tried using this Density = Mass/Volume, where I substituted Denisity= 1.225 and Mass equals 5kg to get the volume as 4.08. Can...

I had already found the Mass of the product (C3H3N) produced by this reaction (theoretical mass at 100% yield) in a previous problem. I did this by finding the Limiting Reagent (C3H6) in the reaction , calculating the number of moles of C3H6 and using the Molar Ratios in the balanced reaction...

I ASSUME THAT THE PRESSES OF THE TWO CONTAINERS WILL BE EQUAL IN THE FINAL (STATIONARY REGIME). SO Pa = Pb naRTa/V = nbRTb/V naTa = nbTb Than , I just need to set up a system My question is , so, will the two pressures at the end be the same? And as for the temperatures, can I also say...

I was puzzling over how to solve this and finally peeked at the solution. They used the relevant equation above. I disagree with this though. The problem specifically says “the piston is allowed to slide freely!” This means that we don’t let it happen slowly. So then we are not in...

P1/V1/T1 = P2 V2 /T2 is derived from the ideal gas equation. However it is stated that this equation breaks down at very high pressures and at very low temperatures. Does anyone know what kind of pressures and temperatures we are talking about here?

Up to an undetermined constant ##a## the entropy of an ideal gas goes like$$S = k_B N\ln \left[ a^\frac{3}{2} \left(\frac{V}{N}\right) \left(\frac{E}{N}\right)^{\frac{3}{2}} \right]$$In some notes is written: And then they identify ##\Omega = \left(\frac{\Delta x \Delta p}{w}\right)^{3N}##...

By using PV=nRT formula, I have found the volume of the vessel. As far as I have learned to calculate the number of collision in a unit volume. So, it is being difficult for me to find the right way to solve. I searched on the internet and have got this...

Here is what I did : work done in going from A to C, W1 = 2nRToln(2) (isothermal process) work done in going from C to B, W1 = pΔV = nRΔT = -nRTo (isobaric process) work done in going from B to A, W3 = 0 (isochoric process) so, total work done = W1 + W2 + W3...

I've first calculated the partial pressures of each gas: ##N_2: 0.4\times 7.4\times 10^4=3.0\times 10^4 Nm^{-2}\\## ##O_2: 0.35\times 7.4\times 10^4=2.6\times 10^4 Nm^{-2}\\## ##CO_2: 0.25\times 7.4\times 10^4=1.9\times 10^4 Nm^{-2}\\## From here, I do not know how to continue. Could someone...

First, I tried using the Archimedes principle and calculated the weight of the surrounding air displaced when taking off. ##W = 2500\times 1.29\times 9.81 = 31637.25 N## But then, I got stuck and do not know how to proceed from here on. I don't want the full solution yet but can I get some...

Attempt at a Solution: Heat Absorbed By The System By the first law of thermodynamics, dU = dQ + dW The system is of fixed volume and therefore mechanically isolated. dW = 0 Therefore dQ = dU The change of energy of the system equals the change of energy of the gas plus the change of energy...

If we assume the energy of particles in an ideal gas follows a Boltzmann distribution, then the energy distribution function can be defined as below: , where k_B is the Boltzmann constant Since the energy of particles in an ideal gas are assumed to only consist of translational kinetic energy...

What do I see in my solution is : ΔW + ΔQ = W_pv + W' + ΔQ (A little difficult to perceive the useful work ) Work on the environment : -p0*(-ΔV) (WHY negative sign?, Is this the work ON the gas?) ΔV=nR (T0/p0 -T/p) By TdS = dQ ΔS + ΔS0 =0 Reversible case: ΔU= -T0ΔS - (-p0(-ΔV)) + W' (WHY...

First of all I thought it was necessary to calculate the temperature(the only data missing for the formula) using the ideal gas equation(since I've already been given 'p' and 'V'), and plug it in the 'v' formula, but the problem immediately occurred when i tried to find out the number of...

First calculate the total pressure, but that gives me p2=p1*t2/t1 = 2,86 bar and partial pressure of 0,87 bar which is wrong. any tips?

It looks more like a computational obstacle, but here we go. Plugging all of these to the partition function: $$Q = \frac{1}{N! h^{3N}} \int -\exp(\frac{1}{2m}(p^2_{r}+p^2_{\phi}/r^2+p^2_{z})+gz)d\Gamma=$$ $$= \frac{1}{N! h^{3N}} \int \exp{(\frac{-1}{2m}p^2_{r})}dp_{r_{1}}...dp_{r_{N}}...

The actual data for the problem and my (and my friend's) attempt at a solution are in the attached file. In a nutshell, this is what happened. I obtained a solution based on the fact that the system is isolated. Thus the initially hot gas moves the partition doing work onto the initially cold...

Since the spherical wave equation is linear, the general solution is a summation of all normal modes. To find the particular solution for a given value of i, we can try using the method of separation of variables. $$ ψ(r,t)=R(r)T(t)ψ(r,t)=R(r)T(t) $$ Plug this separable solution into the...

This is a question in my midterm. I calculated for the answer as c) 11.7 atm by the Ideal Gas Law. The professor states that "all the air is originally at 1 atm" in the prompt indicates an idea of "both 70 L of air and existing 6 L of air in the tank are at 1 atm", and he grades d) 12.7 atm as...

Summary: U=3/2*n*R*T Can some of you help me with this The total internal energy of an ideal gas is 3770 J. If there are 3 moles of the gas at 1 atm, what is the temperature of the gas? I use U=3/2*n*R*T but get the wrong answer, (101 K) but it should be 303 K [Moderator's note: Moved from...

Hello guys. I really try to understand the problem, can you please help.

I am creating a two-dimensional model of an ideal gas, and I was wondering how I should determine initial velocity. Ideally, I would like for the simulation to reach a point where the velocity distribution resembles that of the maxwell-boltzmann curve — will this be achieved if I, say, assign...

Hi, so I found this on another old "AP" High School Finals Exam. I think I may be super lost. Because the only way that I can think about is KE = 3/2kT. And then that the difference of the Kinetic Energy of the air Particles is the Q supplied by the heater inside the air dryer. So ## \frac...

2.1 * 10^-4m/3 Temperature 310K Pressure: 5.3 * 105 Pa So the Ideal gas formula is PV = nRT 2.1*10^-4m^3 Times 5.3*105Pa = n * Gas Constant * Temperature 2.1*10^-4m^3 (*) 5.3*105Pa = # of moles * I'm not sure what I was doing, but the whole equation stuff got hard and I stopped. I left...

I have a compressed pure gas at a specific temperature and volume. (T1, V1) It suddenly (adiabatically) expands until it's at ambient pressure and a specific temperature. (P2, T2). Given: T1, V1, T2, and P2, I want to find P1 and V2. There's a great example in wikipedia which is almost...

Hi All, When dealing with the kinetic theory of gases in thermodynamics, we obtain the result that the mean kinetic energy per atom is (3/2) kT. In considering different samples like 200g of liquid water or a solid cube of lead with one cubic meter, does kT still play an important role in...

Please refer to diagram. V1 is open initially then V2 is open for 5 minutes for pressure to equalize. V1 and V2 are then shut. V3 is opened. What is Vol 2 ? P(final)*V(final) = n(final)* R*T => (Vol1 + Vol2) = n(final)*R*25C/ 0.070 Torr where n(final) = n(Vol1) + n(Vol2) If I shut V3, I...

For the reversible expansion of an ideal gas the heat flowing out of the surroundings and into the system is equal to the work done by the system. Since both system and surroundings have the same constant temperature the entropy increase of the system is equal to the entropy decrease of the...

Attempt: P_1 (initial pressure on the left section) P_2(initial pressure on the right section) T_f, P_f (final pressure for both sections) P_1 (V/3) = N/2 k (3T/2) P_2 (2V/3) = N/2 k (T/2) P_f V/2 = N/2 k T_f Resulting in 4 unknowns and 3 equations... Not enough to find T_f...

I'm having trouble wrapping my head around some thermodynamics and ideal gas law concepts. I don't have a specific textbook question but Just a concept I'm having trouble with. What I'm struggling with is understanding some of the relations between pressure, volume and temperature...

I understand that ##\dot m=\rho Q## and ##{\dot m}_{in}= {\dot m}_{out}## . So one can say that ##\rho Q_1 = \rho Q_2##. But I'm not sure if that equation is correct. I don't know if the density remains constant, or the volume flow rate. And then how I'm also supposed to tie a state equation in...

I have the definition of change in internal energy. $$ \Delta U = Q - W $$ I can get the work by $$ W = \int_{V_1}^{V_2} p dV = p \Delta V $$ however the pressure isn't constant so this won't do. ## W ## is work done by the gas and ## Q ## is amount of heat energy brought into the system. I'm...

Ok, i am struggling to figure something out. I don't know why math is so much easier than physics haha. ok, here is my struggle. I have two states, state 1 and 2, which i will call just 1 and 2. 1: T=298kelvin V=0.025m(cubed) P=310Kpa Mass1=Mass2 R=0.2870 2: T=323kelvin V=0.025m(cubed) P=...

Homework Statement After a free expansion to quadruple its volume, a mole of ideal diatomic gas is compressed back to its original volume isobarically and then cooled down to its original temperature. What is the minimum heat removed from the gas in the final step to restoring its state...

Homework Statement Homework Equations Ideal gas law The Attempt at a Solution The solution to this problem assumes the pressure inside the balloon is the same as the outside pressure, i.e. atmospheric pressure. Is this a valid assumption? I would guess otherwise.

Homework Statement What is a real-life example of the ideal gas law? Homework Equations PV = nRT (Pressure x volume = number of moles x the gas constant x temperature in Kelvin) The Attempt at a Solution https://www.reference.com/science/ideal-gas-law-used-everyday-life-3dacbd6ebd3b5949...

If considering the free expansion of an ideal gas adibatically such that the final volume is double of the initial volume. Since dU=0 for free expansion ,which implies Ti=Tf for ideal gas...(1) From adiabatic relation , TiViγ-1=TfVfγ-1, to satisfy (1) ,γ-1=0 and for ideal gas Cp-Cv=nR or...

Homework Statement An ideal gas has a molar mass of 40 g and a density of 1.2 kg m-3 at 80°C. What is its pressure at that temperature? Homework Equations PV=nRT R constant= 8.314 n= number of moles T= tempreture in kelvin density=Mass/ Volume The Attempt at a Solution i simply solved it like...

Hi everyone, I'd really appreciate any help with this problem: A helium cylinder for the inflation of party balloons hold s 25.0L of gas and is filled to a pressure of 16500kPa at 15 degrees celsius. How many balloons can be inflated from a single cylinder at 30 degrees celsius if the volume of...

The ans comes out (c) if I take specific heat at constt volume to be independent of temp. Whether the specific heat is always temp. independent for an ideal gas??

Homework Statement In all calculations, take R = 8.31 J/m-K. Use the Sackur-Tetrode equation with N replaced by n and K replaced by R to calculate the changes in entropy. Also, assume that these processes are quasi-static so that the ideal gas law and the first law apply at all times. Consider...

What is difference between perfect gas and ideal gas?

The MB energy distribution is: MB_PDF(E, T) = 2*sqrt(E/pi) * 1/(kB*T)^(3/2) * e^(-E/(kB*T)) How do I arrive at the density of states which hides inside the expression 2*sqrt(E/pi) * 1/(kB*T)^(3/2) ? I've only seen DOS that have "h" in them.. I want it to contain only E, pi, kB and T.. This is...

Homework Statement The following is the equation of ideal gas law, where p is pressure (Force/Area), V is volume, n is number of moles and T is temperature in Kelvin. What is the fundamental unit of R? pV = nRT A. kg^−1 · m^−2 · s^ 2 · K · mol B. kg^−1 · m^−4 · s ^2 · K · mol C. kg · m^4 · s...

When an ideal gas,in a piston kind of system and whose equilibrium state is mentioned, is allowed to expand (piston is allowed to move and not gas leaking )against a constant external pressure very quickly, then, is the work done by gas zero or not zero ? The argument for work being zero is...

Homework Statement Basically the units of the Canonical Partition Function within the logarithms should be zero Homework Equations The Attempt at a Solution N here is a number so we ignore the left logarithms, applying a "Unit function " for the terms within the logarithm...