Question 513: Thermodynamics & KTG

Question

10 mole of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The change in the internal energy of the gas is $\_\_\_\_$ cal. (The molecular specific heat of oxygen at constant pressure, $C_P=7 \mathrm{cal} / \mathrm{mol} .{ }^{\circ} \mathrm{C}$ and $\left.\mathrm{R}=2 \mathrm{cal} . / \mathrm{mol} .{ }^{\circ} \mathrm{C}.\right)$

Accepted Answer

For an ideal gas, according to Mayer's relation
$$ C_p-C_v=R $$
Where R is the universal gas constant.
$$ C_v=C_p-R $$
Given:
$\mathrm{C}_{\mathrm{p}}=7 \frac{\mathrm{cal}}{\mathrm{mol}^{\circ} \mathrm{C}}$
$\mathrm{R}=2 \frac{\mathrm{cal}}{\mathrm{mol}^{\circ} \mathrm{C}}$
So, the molecular specific heat of oxygen at constant volume is,
$$ \mathrm{C}_{\mathrm{v}}=7-2=5 \frac{\mathrm{cal}}{\mathrm{~mol}^{\circ} \mathrm{C}} $$
Internal energy change is a state function that depends only on the temperature change. For n moles of an ideal gas, it is given by the formula:
$$ \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} $$
The given values are,
Number of moles $\mathrm{n}=10 \mathrm{~mol}$
Initial temperature, $\mathrm{T}_1=30^{\circ} \mathrm{C}$
Final temperature $\mathrm{T}_2=40^{\circ} \mathrm{C}$
So, the change in temperature is, $\Delta \mathrm{T}=\mathrm{T}_2-\mathrm{T}_1=40-30=10^{\circ} \mathrm{C}$
So, the change in internal energy is,
$$ \Delta U=10 	imes 5 	imes 10=500 \mathrm{cal} $$
Therefore, when 10 moles of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ internal energy increase by 500 cal.
Hence, the correct answer is 500 .

Thermodynamics & KTG

Integer / Numeric Answer Type