Question
(a) How much heat is required to raise the temperature of 1000 grams of nitrogen from $-20^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ at constant pressure?(b) How much has the internal energy of the nitrogen increased?(c) How much external work was done?(d) How much heat is required if the volume is kept constant?Take the specific heat at constant volume $c_v=5 \mathrm{cal} / \mathrm{mole}^{\circ} \mathrm{C}$ and $R=2 \mathrm{cal} / \mathrm{mole} \cdot{ }^{\circ} \mathrm{C}$.
Step 1
The molar mass of nitrogen (N₂) is approximately 28 g/mol. \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \, \text{g}}{28 \, \text{g/mol}} \approx 35.71 \, \text{moles} \] Show more…
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In the following take $C_{V}=20.8$ and $C_{P}=29.1 \mathrm{J} \mathrm{mol}^{-1}$ $^{\circ} \mathrm{C}^{-1}$ for nitrogen gas:(a) Three moles of nitrogen at $303.15 \mathrm{K}\left(30^{\circ} \mathrm{C}\right)$ contained in a rigid vessel, is heated to $523.15 \mathrm{K}\left(250^{\circ} \mathrm{C}\right) .$ How much heat is required if the vessel has a negligible heat capacity? If the vessel weighs $100 \mathrm{kg}$ and has a heat capacity of $0.5 \mathrm{kJ} \mathrm{kg}^{-1}$ $^{\circ} \mathrm{C}^{-1}$ how much heat is required?(b) Four moles of nitrogen at $473.15 \mathrm{K}\left(200^{\circ} \mathrm{C}\right)$ is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to $313.15 \mathrm{K}\left(40^{\circ} \mathrm{C}\right)$ if the heat capacity of the piston and cylinder is neglected?
The temperature of $5.00 \mathrm{~kg}$ of $\mathrm{N}_{2}$ gas is raised from $10.0{ }^{\circ} \mathrm{C}$ to $130.0{ }^{\circ} \mathrm{C}$. If this is done at constant volume, find the increase in internal energy $\Delta U$. Alternatively, if the same temperature change now occurs at constant pressure determine both $\Delta V$ and the external work $\Delta W$ done by the gas. For $\mathrm{N}_{2}$ gas, $c_{v}=0.177$ cal $/ \mathrm{g} \cdot{ }^{\circ} \mathrm{C}$ and $c_{p}=0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}$. If the gas is heated at constant volume, then no work is done during the process. In that case $\Delta W=0$, and the First Law tells us that $(\Delta Q)_{v}=\Delta U$. Because $(\Delta Q)_{v}=c_{v} m \Delta T$, $\Delta U=(\Delta Q)_{v}=\left(0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)=106 \mathrm{kcal}=443 \mathrm{~kJ}$ The temperature change is a manifestation of the internal energy change. When the gas is heated $120^{\circ} \mathrm{C}$ at constant pressure, the same change in internal energy occurs. In addition, however, work is done. The First Law then becomes $$(\Delta Q)_{\mathrm{p}}=\Delta U+\Delta W=443 \mathrm{~kJ}+\Delta W$$ But $(\Delta Q)_{\mathrm{p}}=c_{p} m \Delta T=\left(0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)$ $$=149 \mathrm{kcal}=623 \mathrm{~kJ}$$ Hence $\Delta W=(\Delta Q)_{\mathrm{p}}-\Delta U=623 \mathrm{~kJ}-443 \mathrm{~kJ}=180 \mathrm{~kJ}$
For nitrogen gas the values of $C_{v}$ and $C_{p}$ at $25^{\circ} \mathrm{C}$ are $20.8 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ and $29.1 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1},$ respectively.When a sample of nitrogen is heated at constant pressure, what fraction of the energy is used to increase the internal energy of the gas? How is the remainder of the energy used? How much energy is required to raise the temperature of $100.0 \mathrm{g} \mathrm{N}_{2}$ from $25.0^{\circ} \mathrm{C}$ to $85.0^{\circ} \mathrm{C}$ in a vessel having a constant volume?
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