(I expect the answers to be in algebra ignore the numbers on the axes – they don’t mean anything in particular. Medium Solution Verified by Toppr Given that the process is reversible adiabatic. Compute the final volume reached by gas in this case. This exercise will illustrate that U is a function of state, but Q and W are not. The work done when two mole of an ideal gas is compressed from a volume of 5 m 3 to 1 dm 3 at 300 K, under a pressure of 100 kPa is: A +499.9 kJ B 499.9 kJ C 99. Calculate workdone in adiabatic compression of one mole of an ideal gas (monoatomic) from an initial pressure of 1 atm to final pressure of 2 atm. (b) An isochoric decrease in pressure followed by an isobaric expansion Īt each stage, calculate the work done on or by the gas, the heat gained by the gas or lost from the gas, and the increase or decrease of the internal energy of the gas. (a) An isobaric expansion followed by an isochoric decrease in pressure It is simply defined as the ratio of the molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure, Z V Actual / V Ideal. In figure VIII.2, a gas goes from ( P 1, V 1) to ( P 2, V 2) via three different reversible routes: Think about this! Is it coincidence, or must it be so? In the process, the pressure of the gas is continuously changing. This is also equal to the heat that would be lost if the gas were to cool from T1 to T2 at constant volume. Note also that, since R = C P − C Vand C P/C V = γ this can also be written (Note that T 2 < T 1 in this adiabatic expansion.)Ĭompare this with equation 8.3.1 for an isothermal expansion. This is our final answer to this question. This means that to the nearest joule, 255 joules of work are done by the external force to compress the gas. This gives us an answer of 255.36 joules. PS : I tell all this because I struggled with this barely an year ago.\] Substituting these values in, we find that the work done is equal to 28000 pascals times 0.042 cubic meters minus 0.03288 cubic meters. If it was the other way around, then it would appear with ΔU (Can you see why?). In my picture above, W is work done on the system and thus appears along with Q. I Work done by a gas in a heat engine which compresses a spring. Work done on the gas is represented by the shaded area. compressing ideal gas from p 1 to p 2 adiabatically means the invested energy in the gas is negative and equal to. The only difference this gives you is in the position of W. Work done can be calculated from pV-plot when pressure is not constant and changes are in infinite steps (reversible conditions) during compression from initial volume, V i to final volume, V f. The Volume of the gas moved by the piston is V+ (Ax) or V2/3V1 I don't understand where any of this is getting me I'm sorry. Do not grapple with "is it work done by the system" or "work done on the system". I used the ideal gas equation to get the integral into one variable and get the work done on the surroundings as WnRTln (V2/V1), so that must be the total work done by the gas in expanding. Once you have done so, the corresponding signs of Q, W and ΔU will take care of the situation ( as so elaborately said in post #2 ).īy the way, take care in defining W too.
Mind you, this picture is only to help you gauge the formula properly and place the W in the right place. Thus the change in internal energy is given by the above formula. Now, the internal energy can be increased in two ways : you heat the gas up or/and push the piston down. Consider the system to be a gas kept in a cylindrical bowl with a piston on the top. In that case, you must be using the formula Q + W = ΔU. Also my understanding, is that because the work is done to the system, W should be positive. Two mole of ideal gas diatomic gas (C v, m 5 / 2 R) at 3 0 0 K and 5 atm expanded irreversely and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm.
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