A rigid container is filled with 4.0 mol of air with C_V = 2.5R. How much does the internal (thermal)energy of the air change if its temperature rises from $18 ^ { \circ } \mathrm { C }$ to $448 ^ { \circ } \mathrm { C } \text { ? }$ (R = 8.31 J/mol ? K) A)35,800 J B)430 J C)3580 J D)8940 J

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The internal energy of an ideal gas is given by:$$U = \frac{3}{2}nRT$$where:* U is the internal energy in joules* n is the number of moles of gas* R is the ideal gas constant (8.31 J/mol·K)* T is the temperature in kelvinsIn this problem, we are given that:* n = 4.0 mol* Cv = 2.5R* Ti = 18 °C = 291 K* Tf = 448 °C = 721 KWe can use the equation for internal energy to calculate the change in internal energy:$$\Delta U = Uf - Ui = \frac{3}{2}nR(Tf - Ti)$$$$\Delta U = \frac{3}{2}(4.0 	ext{ mol})(8.31 	ext{ J/mol·K})(721 	ext{ K} - 291 	ext{ K})$$$$\Delta U = 35,800 	ext{ J}$$Therefore, the correct answer is A.

A Rigid Container Is Filled with 4 18∘C18 ^ { \circ } \mathrm { C }18∘C

A Rigid Container Is Filled with 4 $18 ^ { \circ } \mathrm { C }$