Question

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

Answer 1

Hint: Start by writing the combined gas law equation i.e. PV = nRT and apply the changes in the pressure and temperature as per the statement given in the question. Divide these two equations and simplify the fraction to solve for the value of initial temperature.

Complete answer:
We know , The combined gas laws relation is given by
PV = nRT……equation.(1)
Where P is the pressure , V is the volume , n is the number of moles, R is universal gas constant and T is the temperature in kelvin.
 Given , That pressure is increased by 0.4% , so the new pressure becomes
$P' = P\left( {1 + \dfrac{{0.4}}{{100}}} \right)$
Also, The temperature rises by $1^\circ C$, therefore new temperature becomes ,
$T' = T + 1$
Substituting all these values the new combined gas equation will be
P’V = nRT’
$P\left( {1 + \dfrac{{0.4}}{{100}}} \right)V = n(T + 1)R \to eqn.2$
Now , dividing eqn.2 by eqn.1, we get
$\dfrac{{P\left( {1 + \dfrac{{0.4}}{{100}}} \right)V}}{{PV}} = \dfrac{{n(T + 1)R}}{{nRT}}$
On further simplification , we get
$ \Rightarrow \left( {1 + \dfrac{{0.4}}{{100}}} \right) = \dfrac{{(T + 1)}}{T}$
Dividing by T in R.H.S, we get
$
   \Rightarrow \left( {1 + \dfrac{{0.4}}{{100}}} \right) = 1 + \dfrac{1}{T} \\
   \Rightarrow \dfrac{{0.4}}{{100}} = \dfrac{1}{T} \\
$
On rearranging the terms , we get
$
   \Rightarrow T = \dfrac{{1000}}{4} \\
   \Rightarrow T = 250^\circ K \\
$
Therefore, the initial temperature was $250^\circ K$.

So , option A is the correct answer.

Note:
Similar questions can be asked for different volume and temperature or pressure relation, for such questions above mentioned method can be used. Students must know all the laws of chemical combination and gases , Also different conditions such as isothermal , isobar, and mole concepts . Attention must be given while dealing with the temperature in gas law as it is in terms of Kelvin (K) and not in Celsius (C).