Question

The change in the magnitude of the volume of an ideal gas when a small additional pressure \[\Delta P\] is applied at a constant temperature is the same as the change when the temperature is reduced by a small quantity \[\Delta T\] at constant pressure. The initial temperature and pressure of the gas were \[300K\] and \[2\,atm\] respectively. If \[|\Delta T| = {\rm{ }}C|\Delta P|\] then value of C in (K/atm.) is _________

Answer 1

Hint: Here, in the above question it has been asked to find value of C in terms of K/atm, and some most important concepts are used which are very basic like ideal gas equation when temperature is constant also when pressure is constant, but there is difference between both equations differentiating type so take care of that and solve the question.

Formula used:
\[PV = nRT\]
Here,
$P$ = pressure of the gas
$R$= Gas constant
$T$= Temperature of the gas
$n$= number of gas molecule per unit volume

Complete step by step solution:
Let us consider two cases;
Case I: At constant temperature, \[T = {\rm{constant}}\]
We know the ideal gas equation that,
\[PV = nRT\]
Here in above equation if temperature is constant then the whole equation become constant such that:
\[PV = {\rm{constant}}\]
Now, differentiating above equation we get
\[P\Delta V + V\Delta P = 0 \]
\[\Rightarrow \Delta V = - \dfrac{{\Delta P}}{P} \times V\]… \[eq\left( 1 \right)\]

Case II: At constant pressure, \[P = {\rm{constant}}\]
We know the ideal gas equation that,
\[PV = nRT\]
We have to follow the steps from above Case I, we get
\[P\Delta V = nR\Delta T\]
But as it is written in the above question that the temperature is decreasing therefore above equation can be written as:
\[P\Delta V = - nR\Delta T\]...........(- sign indicates decrease)
\[ \Rightarrow \Delta V = \dfrac{{ - nR\Delta T}}{P}\] … \[eq(2)\]

Equating \[eq\left( 1 \right)\]and \[eq(2)\] we get
\[ \Rightarrow - \dfrac{{\Delta P}}{P} \times V = \dfrac{{ - nR\Delta T}}{P}\]
\[ \Rightarrow \Delta T = \Delta P\left( {\dfrac{V}{{nR}}} \right)\]
But according to given condition that, \[|\Delta T| = {\rm{ }}C|\Delta P|\], the above equation can be written as:
\[ \Rightarrow C\Delta P = \Delta P\left( {\dfrac{V}{{nR}}} \right)\]
\[ \Rightarrow C = \left( {\dfrac{V}{{nR}}} \right)\]

Now, we know the ideal gas equation as \[PV = nRT\]
Therefore, \[\left( {\dfrac{V}{{nR}}} \right) = \left( {\dfrac{T}{P}} \right)\]
Thus, after replacing this term in above equation we get,
\[ \Rightarrow C = \left( {\dfrac{T}{P}} \right)\]… \[eq(3)\]
Now, in the question the given data is
\[T = 300K,P = 2atm\]
Therefore, after placing these values in \[eq(3)\], we get
\[ \Rightarrow C = \left( {\dfrac{{300}}{2}} \right)\]
\[\therefore C = 150K/atm\]

Hence, the value of C in (K/atm.) is 150K/atm.

Note: Just read the question thoroughly while attempting there are so many simple concepts used over here, a deep knowledge of thermodynamics is very important like when there is decrease in any quantity we have to use negative signs and use given data carefully.