A sample consisting of 2,0 moles of diatomic perfect gas molecules at 250K is compressed reversibly and adiabatically until its temperature reaches 300K. Given Cv,m 27,5 JK/mol calculate q, w, dU, dH,dS
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To calculate the values of q, w, ΔU, ΔH, and ΔS for the given scenario, we will use the following formulas:
1. q = 0 (since the process is adiabatic)
2. w = -ΔU
3. ΔU = Cv,m * n * ΔT
4. ΔH = ΔU + w
5. ΔS = n * Cv,m * ln(T2/T1)
Given values:
n = 2.0 moles
Cv,m = 27.5 JK/mol
T1 = 250K
T2 = 300K
Step 1: Calculate ΔU
ΔT = T2 - T1
ΔU = Cv,m * n * ΔT
= 27.5 JK/mol * 2.0 mol * (300K - 250K)
= 550 JK
Step 2: Calculate w
w = -ΔU
= -550 JK
Step 3: Calculate ΔH
ΔH = ΔU + w
= 550 JK - 550 JK
= 0 JK
Step 4: Calculate ΔS
ΔS = n * Cv,m * ln(T2/T1)
= 2.0 mol * 27.5 JK/mol * ln(300K/250K)
≈ 19.43 JK
Therefore, the values are:
q = 0 JK
w = -550 JK
ΔU = 550 JK
ΔH = 0 JK
ΔS ≈ 19.43 JK
To calculate q, w, ΔU, ΔH, and ΔS for the given scenario, we can use the first and second laws of thermodynamics. Let's break down each parameter:
1. Heat (q): The process is adiabatic, which means it occurs without the transfer of heat. So, q = 0.
2. Work (w): The work done during an adiabatic process can be calculated using the equation:
w = Cv,m * (T2 - T1)
where:
- Cv,m is the molar heat capacity at constant volume (27.5 JK/mol),
- T1 is the initial temperature (250K),
- T2 is the final temperature (300K).
Plugging in the values, we get:
w = 27.5 JK/mol * (300K - 250K)
3. Change in internal energy (ΔU): The first law of thermodynamics states that ΔU = q - w. As q = 0,ΔU = -w (since heat transfer is zero).
So, ΔU = -w = -27.5 JK/mol * (300K - 250K)
4. Change in enthalpy (ΔH): For an adiabatic process, ΔH = ΔU, since no heat is transferred.
So, ΔH = -27.5 JK/mol * (300K - 250K)
5. Change in entropy (ΔS): The change in entropy can be calculated using the equation:
ΔS = Cv,m * ln(T2/T1)
where:
- Cv,m is the molar heat capacity at constant volume (27.5 JK/mol),
- T1 is the initial temperature (250K),
- T2 is the final temperature (300K).
Plugging in the values, we get:
ΔS = 27.5 JK/mol * ln(300K/250K)
Now you can calculate q, w, ΔU, ΔH, and ΔS using the above formulas.