Calculate the change in entropy (in J/K) when 61.4 g of nitrogen gas is heated at a constant pressure of 1.50 atm from 25.8 ºC to 92.8 ºC. (The molar specific heats are Cv is 20.8 J/(mol-K) and Cp is 29.1 J/(mol-K) .)
To calculate the change in entropy, we need to use the equation:
ΔS = ∫(Cp/T)dT
Where:
- ΔS represents the change in entropy
- Cp is the molar specific heat at constant pressure
- T is the temperature in Kelvin
To solve this problem, we need to convert the given values to Kelvin.
Given:
- Mass of nitrogen gas (m) = 61.4 g
- Pressure (P) = 1.50 atm
- Initial temperature (T1) = 25.8 ºC
- Final temperature (T2) = 92.8 ºC
- Cp = 29.1 J/(mol-K)
First, let's calculate the number of moles (n) of nitrogen gas:
- Convert mass to moles using the molar mass of nitrogen gas (N2), which is 28.0134 g/mol.
n = m / M
n = 61.4 g / 28.0134 g/mol
n ≈ 2.19 mol
Next, we need to calculate the temperature in Kelvin:
- Convert the initial and final temperatures to Kelvin by adding 273.15 to each given temperature.
T1 = 25.8 ºC + 273.15 K ≈ 298.95 K
T2 = 92.8 ºC + 273.15 K ≈ 366.95 K
Now, we can calculate the change in entropy:
ΔS = ∫(Cp / T) dT
ΔS = ∫(29.1 J/(mol-K) / T) dT
To integrate, we need to express Cp/T as a function of T.
ΔS = ∫(29.1 / T) dT
Integrating the expression:
ΔS = 29.1 ln(T) + C
To find C, we substitute the initial temperature (T1) and entropy (S1):
ΔS = 29.1 ln(T2) - 29.1 ln(T1)
Now, we substitute the values (remember to use Kelvin values):
ΔS = 29.1 ln(366.95 K) - 29.1 ln(298.95 K)
ΔS ≈ 29.1(5.907) - 29.1(5.700)
ΔS ≈ 172.08 - 165.87
ΔS ≈ 6.21 J/K
Therefore, the change in entropy when 61.4 g of nitrogen gas is heated is approximately 6.21 J/K.