Question

Ideal gas of $4.48{\text{ L}}$ at STP required $12{\text{ cal}}$ to raise the temperature by ${15^ \circ }{\text{C}}$ at constant volume. The ${{\text{C}}_{\text{P}}}$ of the gas is _____ cal.

• (A) 2

(B) 4

(C) 6

(D) 8

Answer 1

Hint: We have to calculate the heat capacity at constant pressure i.e. ${{\text{C}}_{\text{P}}}$. To solve this we must know the equation which gives the relationship between the heat of the reaction and the heat capacity at constant volume. Calculate the heat capacity at constant volume. Then calculate the heat capacity at constant pressure using the value of heat capacity at constant volume.

Formulae Used:
1. ${\text{q}} = {\text{n}}{{\text{C}}_{\text{V}}}\Delta {\text{T}}$
2. ${{\text{C}}_{\text{P}}} = {{\text{C}}_{\text{V}}} + {\text{R}}$

Complete step-by-step answer:
We know that one mole of an ideal gas at STP occupies a volume of $22.4{\text{ L}}$. STP means standard condition of temperature and pressure. Standard condition of temperature means that the temperature is ${0^ \circ }{\text{C}}$ or $273{\text{ K}}$ and the standard condition of pressure means that the pressure is $1{\text{ atm}}$.
We are given $4.48{\text{ L}}$ of an ideal gas at STP. Thus, we have to calculate the number of moles of ideal gas that occupies a volume of $4.48{\text{ L}}$.
${\text{n}} = 4.48{\text{ L}} \times \dfrac{{1{\text{ mol}}}}{{22.4{\text{ L}}}}$
${\text{n}} = 0.2{\text{ mol}}$
Thus, $4.48{\text{ L}}$ of an ideal gas at STP contains $0.2{\text{ mol}}$ of an ideal gas.
The equation which gives the relationship between the heat of the reaction and the heat capacity at constant volume is as follows:
${\text{q}} = {\text{n}}{{\text{C}}_{\text{V}}}\Delta {\text{T}}$
Where ${\text{q}}$ is the heat of the reaction,
${\text{n}}$ is the number of moles of gas,
${{\text{C}}_{\text{V}}}$ is the heat capacity at constant volume,
 $\Delta {\text{T}}$ is the change in temperature.
Rearrange the equation for the heat capacity at constant volume as follows:
${{\text{C}}_{\text{V}}} = \dfrac{{\text{q}}}{{{\text{n}}\Delta {\text{T}}}}$
Substitute $12{\text{ cal}}$ for the heat of the reaction, $0.2{\text{ mol}}$ for the number of moles of gas and ${15^ \circ }{\text{C}}$ for the change in temperature. Thus,
${{\text{C}}_{\text{V}}} = \dfrac{{12{\text{ cal}}}}{{0.2{\text{ mol}} \times {{15}^ \circ }{\text{C}}}}$
${{\text{C}}_{\text{V}}} = 4{\text{ cal/mol}}{{\text{ }}^ \circ }{\text{C}}$
Thus, the heat capacity at constant volume is $4{\text{ cal/mol}}{{\text{ }}^ \circ }{\text{C}}$.
The equation that gives the relationship between heat capacity at constant volume and heat capacity at constant pressure is as follows:
${{\text{C}}_{\text{P}}} = {{\text{C}}_{\text{V}}} + {\text{R}}$
Where ${{\text{C}}_{\text{P}}}$ is the heat capacity at constant pressure,
 ${{\text{C}}_{\text{V}}}$ is the heat capacity at constant volume,
${\text{R}}$ is the universal gas constant.
Substitute $4{\text{ cal/mol}}{{\text{ }}^ \circ }{\text{C}}$ for the heat capacity at constant volume, $4{\text{ cal/mol}}{{\text{ }}^ \circ }{\text{C}}$ for the universal gas constant. Thus,
${{\text{C}}_{\text{P}}} = \left( {4 + 2} \right){\text{ cal/mol}}{{\text{ }}^ \circ }{\text{C}}$
${{\text{C}}_{\text{P}}} = 6{\text{ cal/mol}}{{\text{ }}^ \circ }{\text{C}}$
Thus, the ${{\text{C}}_{\text{P}}}$ of the gas is $6{\text{ cal}}$.

Thus, the correct option is (C), 6.

Note:We know that ${{\text{C}}_{\text{P}}}$ is the heat capacity at constant pressure. ${{\text{C}}_{\text{P}}}$ is the amount of heat absorbed or released by unit mass of a substance with change in temperature at constant pressure. The change in temperature causes a change in the enthalpy of the system. The heat capacity at constant pressure contributes to the work done as well as the change in internal energy.