Question

Heat capacity of a gasHeat capacity C_vis the amount of heat required to raise the temperature of a given mass of gas with constant volume by 1°C, measured in units of cal/deg-mol (calories per degree gram molecular weight).The heat capacity of oxygen depends on its temperature T and satisfies the formulaC_v = 8.27 + 10^(-5) * (26T − 1.87T²)Use Simpson’s Rule to find the average value of C_v and the temperature at which it is attained for20°C ≤ T ≤ 675°C.

Accepted Answer

Identify the function to analyze: the heat capacity function is given by $$C_v(T) = 8.27 + 10$$^{-5}$$ 	imes (26T - 1.87T^2$)$$, where $$T is $$the temperature in degrees Celsius. Recall that the average value of a continuous function $$f(T) on $$the interval $$[a, b] is $$given by the formula $$\displaystyle \bar{f} = \frac{1}{b - a} \int_a^b f(T) \, dT. $$Here, $$a = 20$$ and $$b = 675. $$Use Simpson's Rule to approximate the definite integral $$\int_{20}^{675} C_v(T) \, dT. To do $$this, divide the interval $$[20, 675]$$ into an even number $$n of $$subintervals, calculate the values of $$C_v(T) at $$the endpoints and midpoints, and apply the Simpson's Rule formula:

$$\displaystyle \int_a^b f(T) \, dT \approx \frac{\Delta T}{3} \left[f(T_0$) + 4 \sum_{i=1,3,5,...}^{n-1} f(T_i) + 2 \sum_{i=2,4,6,...}^{n-2} f(T_i) + f(T_n) ight]$$

where $$\Delta T = \frac{b - a}{n}$$ and $$T_i = a + i \Delta T. $$Calculate the average heat capacity $$\bar{C_v} by $$dividing the approximate integral value by $$(675 - 20)$$, i.e., $$\bar{C_v} = \frac{1}{675 - 20} \int_{20}^{675} C_v(T) \, dT. To $$find the temperature at which $$C_v$$ attains this average value, solve the equation $$C_v(T) = \bar{C_v}$$ for $$T$$ within the interval $$[20, 675]. $$This may require algebraic manipulation or numerical methods such as graphing or using a root-finding algorithm.

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Key Concepts

Heat Capacity at Constant Volume (C_v)

Simpson’s Rule for Numerical Integration

Average Value of a Function over an Interval