Question

One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas?

a) 0, 300 K
b) 5.76 J/K, 300 K
c) 0, 1500 K
d) 5.76 J/K, 1500 K

Answer 1

The change in entropy of the gas during a reversible isothermal expansion where it doubles in volume is 5.76 J/K. Since 1500 J of heat are added during the process and it's isothermal, the temperature of the gas remains constant at 300 K.

Step-by-step explanation:

The change in entropy of an ideal gas during a reversible isothermal expansion can be determined by using the formula ΔS = nRln(V2/V1), where n is the number of moles, R is the universal gas constant, and V2/V1 is the ratio of the final volume to the initial volume. In this scenario, the volume doubles (V2/V1 = 2), and we have 1 mole of an ideal gas. So the change in entropy (ΔS) is 1 mol × 8.314 J/(mol·K) × ln(2) = 5.76 J/K.

The temperature of the gas during a reversible isothermal expansion remains constant. If 1500 J of heat are added to the system, and we know the process is isothermal (constant temperature) and reversible, the temperature can be found using the formula Q = nRTln(V2/V1). Given that Q = 1500 J and ΔS = 5.76 J/K, we can rearrange the entropy formula to find T = Q/(nRln(V2/V1)). Plugging in the values, we get T = 1500 J / (1 mol × 8.314 J/(mol·K) × ln(2)) = 300 K.