Question

A system consisting of 20.0 mol of a monoatomic ideal gas is cooled at constant pressure from a volume of 50.0 L to 10.0 L. The initial temperature was 300 K. What is the change in entropy of the gas?

a) 69.66 J/K
b) 138.32 J/K
c) 207.98 J/K
d) 276.64 J/K

Answer 1

207.98 J/K is the change in entropy.

The correct option is c) 207.98 J/K.

Step-by-step explanation:

The change in entropy
`(\(\Delta S\))`for an ideal gas can be calculated using the formula
`\(\Delta S = nC_p \ln\left((T_f)/(T_i)\right)\),` where \(n\) is the number of moles,
`\(C_p\)` is the molar heat capacity at constant pressure,
`\(T_f\)` is the final temperature, and
`\(T_i\)` is the initial temperature.

Firstly, determine the final temperature 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 gas constant (8.314 J/(mol·K)), and T is the temperature. Rearrange the equation to find
`\(T_f\)` at the final volume:
`\(T_f = (P_1V_1T_1)/(V_2)\), where \(P_1\), \(V_1\), and \(T_1\)` are the initial pressure, volume, and temperature.

Substitute the values into the entropy formula:
`\(\Delta S = nC_p \ln\left((T_f)/(T_i)\right)\)`. Since it's a monoatomic ideal gas,
`\(C_p = (5)/(2)R\)` for constant pressure. Plug in the values and calculate.

In this case,
`\(n = 20.0 \, \text{mol}\), \(V_1 = 50.0 \, \text{L}\), \(V_2 = 10.0 \, \text{L}\), \(T_1 = 300 \, \text{K}\), and \(P_1\)` is not given. However, since the process is at constant pressure,
`\(P_1 = P_2\).` The pressure is canceled out in the entropy formula, so it doesn't affect the result. Calculate
`\(T_f\)` and then substitute into the entropy formula to find
`\(\Delta S\).` The answer is approximately 207.98 J/K.

The correct option is c) 207.98 J/K.