A closed container initially holds 0.040 g of He at 100.0^ C . The container is subsequently heated. Ignoring the expansion of the container, determine the temperature at which the internal energy has increased by 12 J .

Step-by-step solution

The number of moles of He gas is given by n = m M = 0.040 4 = 0.01 mol . Helium is a monatomic gas, so its molar heat capacity at constant volume is C_V = 3 2 R . The change in internal energy is given by U = n C_V T . Substituting the given values: 12 = 0.01 3 2 8.314 T T = 12 2 0.01 3 8.314 = 800 8.314 96.2 K Since a change in temperature in Kelvin is equal to the change in Celsius, T = 96.2^ C . The final temperature is T_f = T_i + T = 100.0^ C + 96.2^ C = 196.2^ C 196^ C .