Question

1.98 moles of an ideal monatomic gas initially at 328 k and a pressure of 4 atm is compressed adiabatically using a constant external pressure pext until equilibrium is reached. the temperature of the gas is then 420 k. calculate for this process. give your answer in j/k. round your answer to two decimal places.

Answer 1

The change in entropy (ΔS) for the adiabatic compression of 1.98 moles of an ideal monatomic gas, which goes from 328 K to 420 K, is 6.10 J/K when rounded to two decimal places.

Step-by-step explanation:

To calculate the change in entropy (ΔS) for the adiabatic compression of an ideal monatomic gas, we must first understand that in an adiabatic process, there is no heat exchange with the surroundings, making ΔQ equal to zero. However, this does not mean that the entropy of the system remains constant since the gas is being compressed and reaches a new equilibrium temperature. The change in entropy can be calculated using the following formula for a monatomic ideal gas:

ΔS = nCvln(T2/T1)

Where ΔS is the change in entropy, n is the number of moles, Cv is the molar heat capacity at constant volume for a monatomic ideal gas (which is 3/2 R), T2 is the final temperature, and T1 is the initial temperature of the gas. Plugging in the values:

ΔS = 1.98 moles × (3/2 × 8.314 J/mol·K) × ln(420 K / 328 K)

Now we perform the calculation:

ΔS = 1.98 × 12.471 × ln(1.2805)

ΔS = 1.98 × 12.471 × 0.2474

ΔS = 6.1042 J/K (rounded to two decimal places)

The change in entropy for this adiabatic compression process is 6.10 J/K when rounded to two decimal places.