How to identify isochoric process from the problem's wording?

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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• 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. Mar 5, 2022 at 18:07
• 1 g of He and 3/8 point there. Mar 6, 2022 at 10:00
• 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. Mar 6, 2022 at 12:22
• 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. Mar 6, 2022 at 18:02

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.