@AdvilSell hi Advil. No, it is a basic principle in physics that there is no absolute velocity, so you cannot say if an inertial frame is at rest or not.
The idea of a circular queue is that if the queue fills up new items added to the queue overwrite old items. Circular queues tend to be used when we want to keep only the latest additions to the queue and we don't care if old items get lost.
Suppose we have an array of length 10 that we're using as the queue. If n is the number of items in the queue we add new items using queue[n++] = newitem. Yes?
So when n reaches 10 it gets reset to zero again and the next new item would go into queue[0] and overwrite what was originally there.
The idea is that if the queue gets full we only keep the most recently added 10 items and any older items are lost.
It's called a circular queue because it is as though the queue were arranged in a circle so when we reach the last place in the queue we go back to the front again.
@Abcd there may be applications where you only want to keep the last N items (for some value of N) and you don't care if older items get lost.
For example suppose you're streaming live audio from the Internet on your phone and you pause the playback.
While you pause the playback your phone will be downloading the live stream and temporarily storing it in a buffer so when you unpause you can resume listening from where you paused. Yes?
So what the streaming app will probably do is overwrite the oldest data. That would mean when you unpause you will have lost a chunk of the audio, but at least it means you'll still hear the latest part of the audio.
This is done with a circular buffer. When the rear position reaches the end of the buffer it wraps round and overwrites the data that was originally at the head of the queue.
For an NPN transistor the $I-b$ current flows into the base and out the emitter, and $I_c$ flows into the collector and out the emitter, so $I_e = I_c + I_b$. Yes?
@Zerix if you look at the PNP transistor the current flows are reversed so current flows into the emitter. A small part of that current flows out the base and the rest flows out the collector. So we still have $I_e = I_b + I_c$.
The reason I mention this is that the question is asking you about the relative sizes of $I_e$, $I_b$ and $I_c$.
So take answer (A). Can $I_c$ be the greatest current?
You're overcomplicating this. The question says we start with 18mL of water and end with 0.0306 m^3 of steam. And the change happens at a constant 1 atm of pressure.
Suppose the water is in a piston, and as it boils it pushes the piston out. The piston is pushing against a pressure of 1atm so when its volume changes by $\Delta V$ it does work of $P\Delta V$. Like I say, you are overcomplicating this.
@Zerix yes, because the whole process is isothermal that must mean the work done is equal to the heat absorbed (so the internal energy $U$ remains unchanged).
I can never remember whether work done on the gas is positive or whether work done by the gas is positive. An to make things worse chemists and physicists use different conventions for this.
It's just a matter of doing enough questions that you get the hang of what the answers should look like. Then when you come to the exam you'll know roughly what the answer should look like even if you don't know exactly what it is.
The problem with these questions is that I don't understand what the question means by the irreversible process. You'd need to get one of the other JEE students to go through exactly what the question means.
@Zerix You do this in two steps. First vapourise the two moles (i.e. 38mL) of water to get 38mL of steam at 100°C. The internal energy increases by twice the molar heat of vapourisation. OK so far?
This isn't what actually happens of course, but inernal energy is a state function so as long as we start and finish in the correct states we can choose any path to move between the two states and the internal energy change will be the same.
In the second step we let the steam expand isothermally to the final volume. But in an isothermal expansion the work done is equal to the heat flow i.e. $dW = dQ$, so $dU = 0$.
@Zerix remember that I said we can choose any path between the initial and final states, because the internal energy change doesn't depend on the path?
So I'm choosing a path in which the first step is constant volume. I'm choosing this path specifically because the work is zero and that makes the calculation of $\Delta U$ easy.
Well the various forms of queue will share a lot of methods and member variables. Writing an interface means all the shared stuff is defined in just one place. Otherwise you'd be repeating the shared code in all the different classes.
I want to know the formula (and what does the symbols stand for) for how much work is needed to compress a certain volume of gas?
