Question

If the pressure of the gas contained in a closed vessel is increased by 0.4%, when heated by $1 \circ C$, the initial temperature must be:
$\text{A} \quad 250K$
$\text{B}\quad 250 \circ C$
$\text{C}\quad 2500K$
$\text{D}\quad 25\circ C$

Answer 1

Hint: In such thermodynamics questions, we have to observe that the change is small or not. If it is small, we can use differentiation to get the correct answer. One more important thing is that students should observe the statement carefully. The vessel is closed. It means that gas cannot exchange matter with the surrounding. Hence moles constant and unless given, for a vessel, the volume of gas is also constant.
Formula used: PV=nRT

Complete step-by-step solution:
Since moles and volume are constant, we can rewrite the ideal gas equation as:
P = kT, where ‘k’ is some constant.
Now, as in the question, the pressure change is given in percentage, the best way to solve this is by using the logarithmic method;
Taking log on both sides, we get;
$log(P) = log(k) + log (T)$
Now, differentiating the equation, we get;
$\dfrac{\Delta P}{P} = 0 + \dfrac{\Delta T}{T}$ [ Since k is a constant and differentiation of a constant is zero]
Percentage basically means the change in quantity over the original quantity.
Now, the term $\dfrac{\Delta P}{ P}$ means the percentage change in pressure which is equal to +0.4% and the term $\Delta T$means the change in temperature which is equal to $+1\circ C$.
Hence putting these values in the given equation, we get;
$0.4% = \dfrac{1}{T}$
Or $T = \dfrac{1}{0.4%} = \dfrac{100}{0.4} = 250 K$
Hence the initial temperature is 250K, option A. is correct.

Note: One might wonder why we didn’t take $250 {}^0 C$ or $250 {}^0 F$ as the final temperature. Well, one should always remember that in thermodynamics, in all the expressions, we ought to use the absolute temperature value i.e. temperature in kelvin. Unless specially mentioned, we take the S.I. units only for our calculations.