# （反對）開夜車 4.1

（問：但是，在現今社會，無論是上班，或是讀書，完全不「開夜車」，又好像不切實際。）

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（問：那樣說有意思嗎？我正正是問你，在時間管理失宜，需要「開夜車」時，該如何自處？）

• 只可以間中，不可以經常。

• 日間中途要有小睡。

• 平均而言，你仍必須要有，充足的睡眠。亦即是話，某一天睡少了，必須於在當個星期，還回「睡債」。

• 例如，如果你的充足睡眠是，每天七小時，而你在某一天只睡了六小時的話，你就有義務，在當個星期的另一天，睡多一小時。

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— Me@2019-06-06 08:23:56 PM

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# 追憶逝水年華, 3

In Search of Lost Time, 3 | （反對）開夜車 3.1 | 止蝕 4

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1. 只要每晚睡少四小時，就每晚可以多四小時溫習。

2. 只要每晚可以多四小時溫習，就可以追回之前，落後了的進度。

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— Me@2019-05-09 10:04:55 PM

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# （反對）開夜車 2.5

（問：「只要飛蚊不惡化」？你怎能保證？）

「飛蚊症」只是某個或某些疾病的一個「症」，而不是「疾病」本身。一日不能知道，我飛蚊症的病因，我也不能有力預防，症狀的加劇。

（問：你不是說，長期夜睡少睡，導致你的飛蚊症嗎？）

— Me@2019-04-13 03:34:33 PM

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# （反對）開夜車 2.4

（問：但你剛才說，飛蚊症「會帶來精神困擾」？）

（問：為什麼呢？）

（問：立刻要求商戶更換。新買的電器，必定在保養期內，可以更換。）

（問：如果只是一兩點的話，應該不會太明顯。我會先用它，直到幾年後才再換。

（問：永久有死點？）

（問：那又真的，十分不自在。

（問：「只要飛蚊不惡化」？你怎能保證？）

「飛蚊症」只是某個或某些疾病的一個「症」，而不是「疾病」本身。一日不能知道，我的飛蚊症的病因，我也不能有力預防，症狀的加劇。

（問：你不是說，長期夜睡少睡，導致你的飛蚊症嗎？）

— Me@2019-03-18 04:47:58 PM

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# （反對）開夜車 2.3

Ken Chan 時光機 1.4.2.3

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（問：那有什麼「嚴重的後果」？）

「嚴重後果」中，較為輕微的，我也經歷過，而且可以視為「不可逆轉」的。

（問：「良性飛蚊」？）

「良性飛蚊」的「良性」，並不是說真的有益，而只是指「不會致盲，只會帶來精神困擾」。

(問：如果開始有飛蚊，自己又怎會知道是，「良性」還是「惡性」呢？）

• 第一，你要在未有飛蚊時，防範於未然。

• 第二，萬一仍然出現飛蚊的話，要立刻看眼科醫生，讓他為你檢查，是否視網膜開始脫落。

• 如果不是，就不用擔心。

• 如果是，可以立刻做手術，制止視網膜脫落，保住完整的視力。視網膜開始脫落，而不立刻做手術制止之，有致盲的風險。

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（問：但你剛才說，飛蚊症「會帶來精神困擾」？）

（問：為什麼呢？）

— Me@2019-02-25 04:01:37 PM

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# （反對）開夜車 2.2

Ken Chan 時光機 1.4.2.2

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The light that burns twice as bright burns half as long, …

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Ken Chan 事後回心一想，才醒起：「Nitrogen 又點會有味咖？」（氮氣又何來會有味道呢？）

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— Me@2019-02-03 07:02:42 AM

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# （反對）開夜車 2.1

Ken Chan 時光機 1.4.2.1

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（我暫時不記得，他以下說話的上半句是什麼。）

… 如何不是那樣，我就毋須於，放榜當天的晚上，在家裡哭。唉！還是不說了。

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— Me@2019-01-18 03:47:50 PM

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# Ken Chan 時光機 2.2

— Me@2019-01-06 02:18:47 PM

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# Ken Chan 時光機 2.1

— Me@2018-12-27 03:50:13 PM

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# 迷宮直升機 5.3

Ken Chan 時光機 1.4.3

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「附加數」比「基礎數」而言，艱深非常，大部分人也覺得，十分辛苦。但是，亦正正是因為「附加數艱深非常」，你才會覺得「核心數容易萬分」。

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（問：程度高了？）

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（問：無論是直升機，或者是鳥瞰圖，好像都是犯規？）

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（問：似乎沒有。但是，那又好像十分辛苦。）

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— Me@2018-12-17 07:05:34 PM

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# 迷宮直昇機 5.2

Ken Chan 時光機 1.4.2

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（問：我又不敢說，我一定會滿分。）

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（問：不太敢，因為，那幾乎沒有可能。但是，我明白你的意思。現在要我做回中一的測驗卷，又真的會覺得，一定會高分，可能會滿分。）

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「附加數」比「基礎數」而言，艱深非常，大部分人也覺得十分辛苦。但是，亦正正是因為「附加數艱深非常」，你才會覺得「核心數容易萬分」。

— Me@2018-12-09 09:03:02 PM

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# Ken Chan 時光機 1.4.1

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Ken Chan 說：「你們現在會盤算，在會考中，各科太概會有什麼目標，奪取什麼等級的成績。但是，我當年全部科目，只有一個目標，就是要『攞 full』」。

— Me@2018-11-18 10:06:43 PM

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# Ken Chan 時光機 1.3

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— Me@2018-10-31 09:39:05 AM

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# Ken Chan 時光機 1.2

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Ken Chan 所提及，其中一個技巧是，在物理科，如果新學一個課題，就應該先做，大量該個課題的 MC（多項選擇題）。那樣，你就可以極速釐清，該個課題中的新概念。

— Me@2018-10-20 11:46:04 AM

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# Ken Chan 時光機 1.1

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— Me@2018-10-03 03:59:04 PM

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# The Jacobian of the inverse of a transformation

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

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In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

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Consider a transformation $\mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2)$, which is an one-to-one mapping from unbarred $x^i$‘s to barred $\bar{x}^i$ coordinates, where $i=1, 2$.

By definition, the Jacobian matrix J of $\mathscr{T}$ is

$J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}$

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Now we consider the the inverse of the transformation $\mathscr{T}$:

$\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)$

By definition, the Jacobian matrix $\bar{J}$ of this inverse transformation, $\mathscr{T}^{-1}$, is

$\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}$

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On the other hand, the inverse of Jacobian $J$ of the original transformation $\mathscr{T}$ is

$J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}$

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If $\bar{J} = J^{-1}$, their (1, 1)-elementd should be equation:

$\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)$

Let’s try to prove that.

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Consider equations

$\bar{x}^1 = \bar{x}^1(x^1,x^2)$

$\bar{x}^2 = \bar{x}^2(x^1,x^2)$

Differentiate both sides of each equation with respect to $\bar{x}^1$, we have:

$A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

$B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

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$A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}$

$B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}$

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$D-C:$

$\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}}$,

results

$\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}$

— Me@2018-08-09 09:49:51 PM

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# Chain Rule of Differentiation

Consider the curve $y = f(x)$.

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$\displaystyle{\frac{d}{dx}}$ is an operator, meaning “the slope of the tangent of”. So the expression $\displaystyle{\frac{dy}{dx}}$, meaning $\displaystyle{\frac{d}{dx} (y)}$, is not a fraction.

In order words, it means the slope of the tangent of the curve $y = f(x)$ at a point, such as point $A$ in the graph.

The symbol $dx$ has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$. It means $\Delta x$ as shown in the graph. In other words,

$dx = \Delta x$

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The symbol $dy$ also has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$. It means the vertical distance between the current point $A(x_0, y_0)$, where $y_0 = f(x_0)$, and the point $C$ on the tangent line $y = mx + c$, where $m$ is the slope of the tangent line. In other words,

$dy = m~dx$

or

$\displaystyle{dy = \left[ \left( \frac{d}{dx} \right) y \right] dx}$

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The relationship of $\Delta y$ and $dy$ is that

$\displaystyle{\Delta y = \frac{dy}{dx} \Delta x + \text{higher order terms}}$

$\displaystyle{\Delta y = \frac{dy}{dx} dx + \text{higher order terms}}$

$\Delta y = dy + \text{higher order terms}$

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Similarly, for functions of 2 variables:

$\displaystyle{\Delta f(x,y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \text{higher order terms}}$

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy}$

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For functions of 3 variables:

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz}$

$\displaystyle{\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt}}$

— Me@2018-07-15 09:30:29 PM

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# 多項選擇題 6

Multiple Choices 6

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（問：哪有那麼多的時間？）

— Me@2018-05-22 06:02:40 PM

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# Van der Waals equation 1.2

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

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In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

$V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}$

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

$P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}$

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These are the reasons that

$V_{\text{ideal}} < V_{\text{measured}}$

$P_{\text{ideal}} > P_{\text{measured}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT$

— Me@2018-05-16 07:12:51 PM

~~~

… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

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# Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20

— Physics Stackexchange

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Ideal gas law:

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

$P_{\text{measured}} = P_{\text{real}}$

$P_{\text{measured}} < P_{\text{ideal gas}}$

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

$V_{\text{measured}} > V_{\text{real}}$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $P_{\text{measured}}$ onto the LHS, since $P_{\text{measured}} < P_{\text{ideal}}$, the LHS will be smaller than the RHS:

$P_{\text{measured}} V_{\text{ideal}} < nRT$

So in order to maintain the equality, a correction term to the pressure must be added:

$\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT$

$P_{\text{ideal}} V_{\text{ideal}} = nRT$

If we substitute $V_{\text{measured}}$ onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

$P_{\text{ideal}} V_{\text{measured}} > nRT$

So in order to maintain the equality, a correction term to the pressure must be subtracted:

$P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT$

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In other words,

$V_{\text{measured}} > V_{\text{real}}$

$V_{\text{ideal}} = V_{\text{real}}$

$V_{\text{measured}} > V_{\text{ideal}}$

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

$P_{\text{measured}} = P_{\text{real}}$

but

$V_{\text{measured}} \ne V_{\text{real}}$?

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How come

$V_{\text{real}} = V_{\text{ideal}}$

but

$P_{\text{real}} \ne P_{\text{ideal}}$?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” $V_{\text{real}}$ has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

$V_{\text{real}} \equiv V_{\text{measured}}$

$P_{\text{real}} \equiv P_{\text{measured}}$

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Or even better, avoid the terms $P_{\text{real}}$ and $V_{\text{real}}$ altogether. Instead, just consider the relationship between $(P_{\text{ideal}}, P_{\text{measured}})$ and that between $(V_{\text{ideal}}, V_{\text{measured}})$.

Whether $X_{\text{measured}}$ is bigger or smaller than $X_{\text{ideal}}$ ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

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