Question

The molar specific heat of oxygen at constant pressure ${{C}_{p}}=7.03cal/mo{{l}^{\circ }}C$ and $R=8.13J/mo{{l}^{\circ }}C$ . The amount of heat taken by 5 mol of oxygen when heated at constant volume from ${{10}^{\circ }}C$ to ${{20}^{\circ }}C$ will be approximately
A.25cal
B.50cal
C.250cal
D.500cal

Answer 1

Hint: Think of the very definition of molar specific heat capacity at constant volume and then you will get the equation for the heat absorbed. Also, we are to use Mayer's formula so as to find molar specific heat capacity at constant volume from the given value of ${{C}_{p}}$.

Formula used:
$\Delta Q=\mu {{C}_{v}}\Delta T$
${{C}_{p}}-{{C}_{v}}=R$

Complete answer:
Heat capacity is simply the ratio of the amount of heat energy that is supplied to the substance to the resultant temperature change. Let us represent heat capacity by ‘S’. That is,
$S=\dfrac{\Delta Q}{\Delta T}$ ……………… (1)
If we were to define a constant characteristic of the same substance, independent of its amount, we could divide S by mass of the substance, m in kg, hence we get specific heat capacity (s) of the substance. That is,
$s=\dfrac{S}{m}$
Dividing equation (1) by m gives,
$s=\left( \dfrac{1}{m} \right)\dfrac{\Delta Q}{\Delta T}$
Instead of mass m, if we specified the amount of substance in terms of moles (μ), we get molar specific heat capacity of the substance. That is,
$C=\dfrac{S}{\mu }$
Dividing equation (1) by μ gives,
$C=\left( \dfrac{1}{\mu } \right)\dfrac{\Delta Q}{\Delta T}$ ……………….. (2)
Let us represent molar specific heat capacity of a substance at constant pressure by,${{C}_{p}}$ and molar specific heat capacity of a substance at constant volume by, ${{C}_{v}}$.
We also have a formula that relates ${{C}_{p}}$ and${{C}_{v}}$ , that is the Mayer’s formula given by,
${{C}_{p}}-{{C}_{v}}=R$ …………………….. (3)
In this question, we are to find the amount of heat taken (∆Q) when oxygen of μ=5mol is heated at constant volume from 10℃ to 20℃ (∆T=10℃).
For constant volume, equation (1) becomes,
${{C}_{V}}={{\left( \left( \dfrac{1}{\mu } \right)\dfrac{\Delta Q}{\Delta T} \right)}_{V}}$
$\Delta Q=\mu {{C}_{V}}\Delta T$ ……………………… (4)
From equation (3),
${{C}_{v}}={{C}_{p}}-R$ ……………….. (5)
Substituting equation (5) in (4) gives,
$\Delta Q=\mu ({{C}_{P}}-R)\Delta T$ ……………… (6)
Given values from question,
${{C}_{P}}=7.03cal/mo{{l}^{\circ }}C$
$R=8.13J/mo{{l}^{\circ }}C$
We know that, 1cal=4.2joules,
Therefore, $R=\dfrac{8.13}{4.2}cal/mo{{l}^{\circ }}C$
Substituting the given values from the question in (6),
$\Delta Q=5\times \left( 7.03-\dfrac{8.13}{4.2} \right)\times 10$
$\Delta Q=254.71cal\approx 250cal$

So, the correct answer is “Option C”.

Note:
While reading the question, one should specifically note that the given molar specific heat capacity is that at constant pressure and we are to find the amount of heat absorbed at constant volume. If we fail to notice this, we may not use Mayer’s formula and may end up in the wrong answer.