As I mentioned, I agree with Nisarg Bhavsar's conclusion.
For the justification, I think a more formal one can be obtained starting from this Wikipedia article on mass balance:
https://en.wikipedia.org/wiki/Mass_balance#Ideal_batch_reactor
In this case the system is indeed closed, so the mass (not concentration) balance is:
$r_A \cdot V = \frac {dn_A}{dt}$
The reaction rate $r_A$ by convention is moles of A produced minus transformed, per unit time, per unit volume. [That's why its product by the instant reaction volume $V$ gives indeed the change in moles per unit time].
It is stated that the reaction is first order in $A$.
This implies the following:
$r_A = -k \cdot n_1 \cdot C_A$
The stoichiometric coefficient $n_1$ must be added because, again by convention, the rate constant $k$ is given for the 'standard' case where $n_1 = 1$.
Substituting in the mass balance:
$-k \cdot n_1 \cdot C_A \cdot V = \frac {dn_A}{dt}$
As $C_A \cdot V = n_A$, this becomes:
$-k \cdot n_1 \cdot n_A = \frac {dn_A}{dt}$
which leads to the same conclusion as the one from the book and from Nisarg Bhavsar, i.e.:
$n_A(t) = n_{A,0} \cdot e^{-k \cdot n_1 \cdot t}$
The conversion or advancement is:
$y_A(t) = \frac {n_{A,0}-n_A(t)}{n_{A,0}} = 1 - e^{-k \cdot n_1 \cdot t}$
This is in line with the idea that the conversion of a first order reaction does not depend on the initial concentration.
In practice, if one wanted to follow this reaction experimentally, they could sample it, taking only a very small volume of gas out, and measure the concentration of $A$ in the sample, e.g. by gas chromatography.
Knowing what the total volume of the reaction is at the sampling time, one could calculate the total number of moles of $A$, to fit the above integrated equation.
Could one use 'partial volumes' instead of number of moles?
Yes. Given that it is stated that this reaction is occurring at constant (total) pressure and temperature, and the mixture is ideal:
$V = n \cdot \frac { R \cdot T} {P} = n \cdot constant$
$V_A = n_A \cdot constant$
So one can substitute $n_A$ by $V_A$, and the equation still holds.
Why did Atkins' example about azomethane use partial pressure?
$\ce{CH3-N=N-CH3(g) \rightarrow CH3-CH3(g) + N2(g)}$
If one starts with pure azomethane, this is clearly a reaction (in the gas phase, at 600 K) where the total number of moles of gas are doubled when the reaction is complete.
Atkins states that the reaction is first order in azomethane, and the system is closed.
So the same equation as above applies (this time with stoichiometric coefficient $1$):
$-k \cdot n_A = \frac {dn_A}{dt}$
Atkins uses the following integrated law:
$\ln {\frac p {p_0}} = - k \cdot t$
where $p$ is the partial pressure of azomethane (not sure how it was determined experimentally).
By showing that the $\ln p$ vs $t$ data can be fitted to a straight line, it concludes that the reaction was first order.
However, this requires that the reaction be at constant volume (not mentioned by Atkins, but in fact mentioned in other books, like Silbey, Alberty - Physical Chemistry).
Atkins' argument starts from the assumption that:
$\ln {\frac {[A]} {[A]_0}} = - k \cdot t$
and then it says "the partial pressure of a gas is equivalent to its concentration", thus substituting $[A]$ by $p_A$.
It is true that for an ideal gas mixture:
$p_A = x_A \cdot P = \frac {n_A} {n} \cdot P$
and, as:
$P = n \cdot \frac {R \cdot T} V$
$\implies p_A = n_A \cdot \frac {R \cdot T} V = C_A \cdot {R \cdot T}$
So, as long as the temperature is constant, it is indeed legitimate to substitute $C_A = [A]$ by $p_A$, in a ratio where the constant term ${R \cdot T}$ cancels out.
But this is missing the point.
The real question is: where did the integrated rate law come from in the first place?
Going back to the mass balance:
$-k \cdot C_A \cdot V = \frac {dn_A}{dt}$
As far as I can tell, the only way for this to become the integrated rate law used by Atkins is by assuming constant volume. Then, dividing by $V$ on both sides and taking $V$ into the differential:
$-k \cdot C_A = \frac {d \frac{n_A} V}{dt} = \frac {dC_A}{dt}$
which, integrated, is indeed:
$\ln {\frac {C_A} {C_{A,0}}} = \ln {\frac {[A]} {[A]_0}} = - k \cdot t$
If, like in the example from the original post, the total pressure, not the total volume, were constant, one could not derive this equation.
The one based on the number of moles of course would still be valid, but then:
$n_A(t) = n_{A,0} \cdot e^{-k \cdot t}$
$p_A = \frac {n_A} {n} \cdot P$
$n = n_A + n_{ethane} + n_{N_2} = n_A + 2 \cdot (n_{A,0} - n_A) = 2 \cdot n_{A,0} - n_A$
$p_A = \frac {n_A} {2 \cdot n_{A,0} - n_A} \cdot P$
$n_A = \frac {2 \cdot n_{A,0} \cdot p_A}{p_A + P}$
$\frac {2 \cdot n_{A,0} \cdot p_A(t)}{p_A(t) + P} = n_{A,0} \cdot e^{-k \cdot t}$
$p_A(t) = P \cdot \frac {e^{-k \cdot t}} {2 - e^{-k \cdot t}}$
I tested this out numerically and it seems to work, but then again, I am not a physical chemist, so if anyone sees any errors in my logic or if I am using circular arguments, please do correct me.
On the other hand, if Nisarg Bhavsar and I are correct, maybe someone should warn Jayadithya that the accepted answer is not correct (is it?).