Question

In a constant volume calorimeter, 5 g of gas with molecular weight 40 was burnt in excess of oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.75 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 $kJ{K^{ - 1}}$, the numerical value for the $\Delta U$ of combustion of the gas in $kJmo{l^{ - 1}}$is:
A. 15
B. 12
C. 90
D. 8

Answer 1

Hint: To solve this question, we need to understand the concept behind finding the value of $\Delta U$ by the calorimeter. As we know that, a calorimeter is isolated from the rest of the universe, so we can say that the reactants of the system and the rest of the calorimeter to be the surroundings. Thus, the change in internal energy of the reactants upon combustion can be calculated with the formula $dU = - n{C_v}dT$.

Complete step by step answer:As we know that, the calorimeter is an object which can be used for calorimetry. It is the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.
By referring to the question, we know that for an ideal gas in any process, the formula to be used is:
$\Delta U = \dfrac{{{\text{Heat capacity}} \times {\text{Change in temperature}}}}{{{\text{Number of moles}}}}$
Now, we will note all the given quantities:
Molecular weight of gas= $40g$
Weight of gas= $5g$
So, we can calculate the number of moles= $\dfrac{5}{{40}} = 0.125mole$
Heat capacity of the reaction=$2.5kJ{K^{ - 1}}$
Thus, change in temperature= final temperature – initial temperature=$\left( {298.75 - 298} \right)K = 0.75K$
Now, we will substitute all the given quantities in the above formula.
$\Delta U = \dfrac{{2.5 \times 0.75}}{{0.125}} = 15kJmo{l^{ - 1}}$
Therefore, the numerical value for the $\Delta U$ of combustion of the gas in $kJmo{l^{ - 1}}$is $15kJmo{l^{ - 1}}$

Therefore, the option A is correct.


Note:After solving this question, we also need to understand the principle of calorimetry or the principle of mixtures which states that for an insulated system, the amount of heat energy lost by any hot body is equal to the heat energy gained by any cold body. For which, the formula is:
${m_1}{c_1}({t_1} - t) = {m_2}{c_2}(t - {t_2})$
Also, heat transfer occurs when both the bodies attain the same temperature.