Question

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 °C was compressed suddenly (and irrevers- ibly) to 0.500 L by driving in a piston. In the process, the tempera- ture of the gas increased to 28.1°C. Assume ideal behavior, what is the change in entropy of the gas?

Answer 1

The change in entropy of the gas is approximately 4.57 J/K.

Step-by-step explanation:

The change in entropy of the gas can be calculated using the formula:

ΔS = nCv ln(T₂/T₁) + nR ln(V₂/V₁)

Where:

• ΔS is the change in entropy
• n is the number of moles of gas
• Cv is the molar heat capacity at constant volume
• T₁ and T₂ are the initial and final temperatures, respectively
• V₁ and V₂ are the initial and final volumes, respectively
• R is the ideal gas constant

First, we need to calculate the values for n, Cv, T₁, T₂, V₁, and V₂ using the given information.

n = 3.00 L × (1 mol/22.4 L) = 0.134 mol

T₁= 18.5 °C + 273.15 = 291.65 K

T₂ = 28.1 °C + 273.15 = 301.25 K

V₁ = 3.00 L

V₂ = 0.500 L

Next, we need to calculate Cv using the equation:

Cv = (5/2)R = (5/2)(8.314 J/(mol·K)) = 20.785 J/(mol·K)

Now, we can substitute all the values into the formula to calculate the change in entropy:

ΔS = (0.134 mol)(20.785 J/(mol·K)) ln(301.25 K/291.65 K) + (0.134 mol)(8.314 J/(mol·K)) ln(0.500 L/3.00 L)

ΔS ≈ 4.57 J/K

Answer 2

Answer : The change in molar entropy of the sample is -14.22 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

`\Delta S=n* C_v* \ln ((T_2)/(T_1))+n* R* \ln ((V_2)/(V_1))`

where,

`\Delta S` = change in molar entropy

n = number of moles = 1 mole

`T_2` = final temperature =
`28.1^oC=273+28.1=301.1K`

`T_1` = initial temperature =
`18.5^oC=273+18.5=291.5K`

`V_2` = final volume = 0.500 L

`V_1` = initial volume = 3.00 L

`C_(v)` = heat capacity diatomic gas
`(N_2)` =
`(5)/(2)R`

R = gas constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

`\Delta S=1* ((5)/(2)R)* \ln ((301.1)/(291.5))+1* R* \ln ((0.500)/(3.00))`

`\Delta S=1* ((5)/(2)* 8.314)* \ln ((301.1)/(291.5))+1* 8.314* \ln ((0.500)/(3.00))`

`\Delta S=-14.22J/K.mol`

Therefore, the change in molar entropy of the sample is -14.22 J/K.mol