A sample consisting of 2 moles of diatomic perfect gas molecules at 250K is compressed reversibly and adiabatically until its temperature reaches 300K. Given that Cv,m is 27.5 JK/mol calculate q, w, delta U, delta H and delta S.
To calculate q (heat), w (work), ΔU (change in internal energy), ΔH (change in enthalpy), and ΔS (change in entropy), we need to use the First Law of Thermodynamics and relevant thermodynamic equations.
First, let's calculate q (heat) using the equation:
q = ΔU + w
In this case, since the compression is adiabatic (no heat exchange with the surroundings), q = 0.
Next, let's calculate w (work) using the equation:
w = -PΔV
To find the pressure (P) and volume change (ΔV), we need to apply the ideal gas law:
PV = nRT
Given that the initial volume (V) is unknown, we need to find it using the ideal gas law and the given conditions:
V = nRT / P
V = (2 mol)(8.314 J/(mol·K))(250 K) / P
Now, we can calculate the final volume (V_f) using the same equation:
V_f = nRT / P
V_f = (2 mol)(8.314 J/(mol·K))(300 K) / P
The change in volume (ΔV) is:
ΔV = V_f - V
ΔV = [(2 mol)(8.314 J/(mol·K))(300 K) / P] - [(2 mol)(8.314 J/(mol·K))(250 K) / P]
Now we can calculate work (w):
w = -PΔV
w = -P[(2 mol)(8.314 J/(mol·K))(300 K) / P - (2 mol)(8.314 J/(mol·K))(250 K) / P]
w = -2(8.314 J/(mol·K))(300 K - 250 K)
w = -2(8.314 J/(mol·K))(50 K)
Now that we have calculated work (w), we can calculate ΔU (change in internal energy) using the equation:
ΔU = q - w
Since q = 0 (adiabatic process), ΔU = -w
ΔU = -w = 2(8.314 J/(mol·K))(50 K)
To calculate ΔH (change in enthalpy), we can use the equation:
ΔH = ΔU + PΔV
ΔH = ΔU + PΔV
ΔH = 2(8.314 J/(mol·K))(50 K) + P[(2 mol)(8.314 J/(mol·K))(300 K) / P - (2 mol)(8.314 J/(mol·K))(250 K) / P]
ΔH = 2(8.314 J/(mol·K))(50 K) + 300(2 mol)(8.314 J/(mol·K)) - 250(2 mol)(8.314 J/(mol·K))
Finally, to calculate ΔS (change in entropy), we can use the equation:
ΔS = q_rev / T
Since the process is adiabatic, it is not reversible. Therefore, ΔS cannot be calculated.
To summarize:
q = 0 (adiabatic process)
w = -2(8.314 J/(mol·K))(50 K)
ΔU = -w = 2(8.314 J/(mol·K))(50 K)
ΔH = 2(8.314 J/(mol·K))(50 K) + 300(2 mol)(8.314 J/(mol·K)) - 250(2 mol)(8.314 J/(mol·K))
ΔS = Cannot be calculated (not a reversible process)