Question

Given, \[5\] moles of oxygen are heated at constant volume from \[10^{\circ}C\] to \[20^{\circ}C\]. the change in internal energy of the gas is:
\[{\text{Cp = 7}}{\text{.03calmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\] and \[{\text{R = 8}}{\text{.31Jmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\]
A) \[{\text{125cal}}\]
B) \[{\text{252cal}}\]
C) \[{\text{50cal}}\]
D) \[{\text{500cal}}\]

Answer 1

Hint: We have to know that the moles are the one of the main units in chemistry. The moles of the molecule depend on the mass of the molecule and the molecular mass of the molecule. Chemical reactions are measured by moles only. The number of the equivalent of the reactant is also dependent on the moles of the molecule. The number of moles of the reactant and product is equal in the equilibrium reaction.
Formula used:
Gas constant at calories is calculated,
 The gas constant at calories is equal to the gas constant at joules divided by \[4.188\].
\[{\text{gas constant calories = }}\dfrac{{{\text{gas constant joules}}}}{{{\text{4}}{\text{.188}}}}\]
The constant pressure and constant volume relationship with a gas constant is
\[{{\text{C}}_{\text{P}}}{\text{ - }}{{\text{C}}_{\text{V}}}{\text{ = R}}\]
Here,
Constant pressure is represented as ${C_p}$ .
The constant volume is represented as ${C_v}$ .
The gas constant is represented as $R$ .
We must have to know that the heat absorbed in the reaction is equal to the product of constant volume, change in temperature , and number of moles of molecules involved in absorption.
Formula,
\[{\text{q = n}}{{\text{C}}_{\text{V}}}{\Delta T}\]
The number of moles is represented by $n$
Change in temperature is represented by $\Delta T$ .
Change in internal energy is equal to the sum of heat absorbed and work done by the reaction.
\[\Delta U = w + q\]
Work done is represented by $w$ .
Change in internal energy is represented by $\Delta U$ .

Complete step by step answer:
Gas constant at joules is, \[{\text{R = 8}}{\text{.31Jmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\]
Constant pressure, \[{\text{Cp = 7}}{\text{.03calmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\]
Change in temperature, $\Delta T = 10^{\circ}C$ to \[20^{\circ}C\]
\[ \Rightarrow \Delta T = 20^{\circ}C - 10^{\circ}C\]
\[ \Rightarrow \Delta T = 10^{\circ}C\]
Number of moles, $n = 5$
Gas constant at calories is calculated,
The gas constant at calories is equal to the gas constant at joules divided by \[4.188\].
\[{\text{gas constant calories = }}\dfrac{{{\text{gas constant joules}}}}{{{\text{4}}{\text{.188}}}}\]
\[{\text{gas constant calories = }}\dfrac{{{\text{8}}{\text{.31}}}}{{{\text{4}}{\text{.188}}}}\]
\[{\text{ = 1}}{\text{.99calmo}}{{\text{l}}^{{\text{ - 1}}}}{\text{de}}{{\text{g}}^{{\text{ - 1}}}}\]
The constant pressure and constant volume relationship with a gas constant is
Constant pressure, \[{{\text{C}}_p}{\text{ = 7}}{\text{.03calmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\]
Gas constant at joules is, \[{\text{R = 1}}{\text{.99calmo}}{{\text{l}}^{{\text{ - 1}}}}{\deg ^{{\text{ - 1}}}}\]
\[{{\text{C}}_{\text{P}}}{\text{ - }}{{\text{C}}_{\text{V}}}{\text{ = R}}\]
We can change the formula our need,
\[{{\text{C}}_{\text{V}}} = {{\text{C}}_{\text{P}}}{\text{ - R}}\]
Substitute the known values in the formula,
\[{{\text{C}}_{\text{V}}} = {{\text{C}}_{\text{P}}}{\text{ - R}}\]
Now we can substitute the known values we get,
\[{{\text{C}}_v}{\text{ = 7}}{\text{.03 - 1}}{\text{.99}}\]
On simplification we get,
\[{{\text{C}}_v}{\text{ = 5}}{\text{.04calmo}}{{\text{l}}^{{\text{ - 1}}}}{\text{de}}{{\text{g}}^{{\text{ - 1}}}}\]
\[5\] moles of oxygen are heated absorbed from \[10^{\circ}C\] to \[20^{\circ}C\] is
\[{\text{q = n}}{{\text{C}}_{\text{V}}}{\Delta T}\]
Now we can substitute the known values we get,
\[q = 5 \times 10 \times 5.04\]
On simplification we get,
\[ \Rightarrow {\text{q = 252cal}}\]
Here we know that no external work was done at constant volume because of no change in volume in the system.
Therefore, the value of w is zero.
The change in internal energy will be equal to the heat absorbed in the system because of no change in volume hence no work is done in the system.
\[ \Delta U = w + q\]
Now we can substitute the known values we get,
\[\Delta {\text{U = 252 + 0}}\]
On simplification we get,
\[\Delta {\text{U = 252cal}}\]
Hence, the change in internal energy is \[{\text{252cal}}\].
\[5\] moles of oxygen are heated at constant volume from \[10^{\circ}C\] to\[20^{\circ}C\].
The change in internal energy of the gas is \[{\text{252cal}}\] , if the value of \[{\text{Cp = 7}}{\text{.03calmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\] and \[{\text{R = 8}}{\text{.31Jmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}\]

Hence option B is correct.

Note:
We have to remember that the number of moles of the reaction if calculated, in this way we can predict the reaction is not. The number of moles is used to find the yield percentage of the reaction. The moles are also used to find the formation of side products in the chemical reaction. Carbon dioxide is one of the gases used for cooling the reaction. If one reaction carbon dioxide evolves means that reaction is considered an endothermic reaction.