A sealed rigitd tank contains 30 moles of an ideal gas, at an initial temperature of $270 \mathrm {~K}$ The pressure of the gas is increased until the final pressure equals 1.40 times the initial pressure. The heat capacity at constant pressure of the gas is $32.0 \mathrm {~J} / \mathrm { mol } \cdot \mathrm { K }$ What is the change in the internal (thermal)energy of the gas during this process? (R = 8.31 J/mol ? K)
A)77 kJ
B)100 kJ
C)130 kJ
D)-50 kJ
E)-23 kJ

Question

Quizplus · Accepted Answer

The change in internal energy is given by:$$\Delta U = nC_p \Delta T$$where n is the number of moles of gas, $C_p$ is the heat capacity at constant pressure, and $\Delta T$ is the change in temperature.We can find $\Delta T$ using the ideal gas law:$$PV = nRT$$where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.Since the volume is constant, we have:$$\frac{P_i}{T_i} = \frac{P_f}{T_f}$$where $P_i$ and $T_i$ are the initial pressure and temperature, and $P_f$ and $T_f$ are the final pressure and temperature.Solving for $T_f$, we get:$$T_f = T_i \frac{P_f}{P_i}$$Substituting this into the equation for $\Delta U$, we get:$$\Delta U = nC_p \left(T_i \frac{P_f}{P_i} - T_i\right)$$$$= nC_p T_i \left(\frac{P_f}{P_i} - 1\right)$$$$= 30 \mathrm{~mol} \times 32.0 \mathrm{~J} / \mathrm{~mol} \cdot \mathrm{~K} \times 270 \mathrm{~K} \left(\frac{1.40}{1} - 1\right)$$$$= 77 \mathrm{~kJ}$$Therefore, the change in internal energy is 77 kJ.

A Sealed Rigitd Tank Contains 30 Moles of an Ideal 270 K270 \mathrm {~K}270 K

A Sealed Rigitd Tank Contains 30 Moles of an Ideal $270 \mathrm {~K}$