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Problem 19 from NEET's Solved Paper 2013:

The amount of heat energy required to raise the temperature of $\pu{1 g}$ of helium at NTP from $T_1$ to $T_2$ (in kelvin) is

\begin{align} &\text{(a)}~\displaystyle\frac 3 8 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1) &\quad &\text{(b)}~\displaystyle\frac 3 2 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1) \\ &\text{(c)}~\displaystyle\frac 3 4 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1) &\quad &\text{(d)}~\displaystyle\frac 3 4 N_\mathrm{A}k_\mathrm{B}\frac{T_2}{T_1} \end{align}

My chemistry teacher said the “language of question” gives away the process is isochoric and suggested to use the following equation to solve the problem:

$$\mathrm dQ = nC_V\Delta T = \frac f 2nR\Delta T$$

yielding the correct answer $\text{(a)}~\displaystyle\frac 3 8 N_\mathrm{A}k_\mathrm{B}(T_2 - T_1).$

How can this be deduced conceptually? What exactly points to an isochoric process in this case?

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    $\begingroup$ Welcome to CH SE site! Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. Suitable formatting can be done according to formatting math/chem expressions/equations. $\endgroup$
    – Poutnik
    Commented Mar 5, 2022 at 17:50
  • $\begingroup$ Search for molar heat capacity of ideal gases at constant p ( isobaric process ) and constant V ( isochoric process ) // Look also at molar versus specific heat capacity. // Generally, search properly before asking. The site tries to avoid answering questions for which answers can be easily found. $\endgroup$
    – Poutnik
    Commented Mar 5, 2022 at 18:07
  • $\begingroup$ 1 g of He and 3/8 point there. $\endgroup$
    – Poutnik
    Commented Mar 6, 2022 at 10:00
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    $\begingroup$ I guess the implication of the problem statement, especially NTP, suggests (slightly) that the process is at constant pressure. But it certainly is not clear. $\endgroup$ Commented Mar 6, 2022 at 12:22
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    $\begingroup$ Note that NTP cannot be assumed to be at constant p, as it does not assume constant T either. My interpretation is NTP is meant as initial conditions. $\endgroup$
    – Poutnik
    Commented Mar 6, 2022 at 18:02

1 Answer 1

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The heat given to a system can be utilised in increasing its internal energy or doing work against the surroundings. This is popularly known as first law of thermodynamics.
The change in the internal energy brings the change in the temperature and vise versa. So to increase the temperature one should increase the internal energy of the system. We know that, $$ \Delta U = nC_v \Delta T $$ According to first law of thermodynamics, the heat given to the system at constant volume is known as internal energy. ( $\because$ work done ($W$) = 0 at constant volume )$$ \Delta Q = \Delta U $$ $$\implies \Delta Q = nC_v\Delta T $$ I think this clarifies your doubt.

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    $\begingroup$ This avoids the major question why isochoric and not isobaric. :-) $\endgroup$
    – Poutnik
    Commented Mar 6, 2022 at 16:05
  • $\begingroup$ In question, they had mentioned 1 gram of Helium. So, now the moles of helium is fixed, and the temperature is changing. Thus, now there are two possibilities, that the volume can change or the pressure will change. Now, how we will decide that what will change ? @Poutnik $\endgroup$ Commented Mar 7, 2022 at 12:29
  • $\begingroup$ @UjjawalMishra Reread all what has been posted, including comments. It is there. Molar heat capacity of monoatomic ideal gas is (3/2)R for isochoric process, (5/2)R for isobaric one. $\endgroup$
    – Poutnik
    Commented Mar 7, 2022 at 12:31
  • $\begingroup$ It should be pointed out that ∆U=ncv∆T is true for isochoric processes with real substances (assuming constant heat capacity), any process with an ideal gas, but not generally true for any process with any substance. $\endgroup$ Commented Mar 10, 2022 at 7:35
  • $\begingroup$ No you are misconcepted. It is true for all processes. $\endgroup$
    – Infinite
    Commented Mar 10, 2022 at 13:21

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