Question

The specific heat at constant volume for an ideal gas is given by \(c_{v}=0.7+\left(2.7 \times 10^{-4}\right) T(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K})\) where \(T\) is in kelvin. The change in the internal energy for this ideal gas undergoing a process in which the temperature changes from 27 to \(127^{\circ} \mathrm{C}\) is most nearly \((a) 70 \mathrm{kJ} / \mathrm{kg}\) \((b) 72.1 \mathrm{kJ} / \mathrm{kg}\) \((c) 79.5 \mathrm{kJ} / \mathrm{kg}\) \((d) 82.1 \mathrm{kJ} / \mathrm{kg}\) \((e) 84.0 \mathrm{kJ} / \mathrm{kg}\)

Short answer

Answer: The change in internal energy is approximately 79.5 kJ/kg.

Step-by-step solution

Convert temperatures to Kelvin

First, convert the given initial and final temperatures from Celsius to Kelvin by adding 273.15 to both temperatures: Initial temperature, \(T_1 = 27^\circ C + 273.15 K = 300.15 K\) Final temperature, \(T_2 = 127^\circ C + 273.15 K = 400.15 K\)

Find specific heat at both temperatures

Use the given formula for specific heat at constant volume \(c_v = 0.7 + (2.7 \times 10^{-4}) T\) to find \(c_{v_1}\) and \(c_{v_2}\): \(c_{v_1} = 0.7 + (2.7 \times 10^{-4}) (300.15) = 0.7 + 0.0810405 = 0.7810405 \frac{kJ}{kg \cdot K}\) \(c_{v_2} = 0.7 + (2.7 \times 10^{-4}) (400.15) = 0.7 + 0.1080405 = 0.8080405 \frac{kJ}{kg \cdot K}\)

Calculate average specific heat

Now, calculate the average specific heat during the process by taking the arithmetic mean of the specific heat at the initial and final temperatures: \(c_{v_{avg}} = \frac{c_{v_1} + c_{v_2}}{2} = \frac{0.7810405 + 0.8080405}{2} = 0.7945405 \frac{kJ}{kg \cdot K}\)

Calculate change in internal energy

Use the formula for the change in internal energy \(\Delta U = mc_{v_{avg}}(T_2 - T_1)\), where \(m\) is the mass of gas. Since we need the answer in \(\frac{kJ}{kg}\), we can set \(m\) as 1 to get the desired units: \(\Delta U = 1 \cdot 0.7945405 (400.15 - 300.15) = 0.7945405 \cdot 100 = 79.45405 \frac{kJ}{kg}\)

Compare to given options

Now, compare 79.45405 to the given options and choose the one nearest to it: (a) \(70 \frac{kJ}{kg}\) (b) \(72.1 \frac{kJ}{kg}\) (c) \(79.5 \frac{kJ}{kg}\) (d) \(82.1 \frac{kJ}{kg}\) (e) \(84.0 \frac{kJ}{kg}\) The most nearly correct answer is (c) \(79.5 \frac{kJ}{kg}\).