Answer:
P1 = 45,174 Pa
P2 = 90,348 Pa
W = 2,259 J
Q = 2,259 J
ΔS = 0
Step-by-step explanation:
We can use the ideal gas law, PV = nRT, to solve this problem. Since the gas is at constant temperature (isothermal), we can simplify this to PV = constant.
Given that there are two moles of oxygen gas in a volume of 0.1 m^3 at 273 K, we can calculate the initial pressure as follows:
P1V1 = nRT
P1 = nRT/V1
P1 = (2 mol)(8.31 J/mol.K)(273 K)/(0.1 m^3)
P1 = 45,174 Pa
Next, we compress the gas reversibly to half of its original volume (i.e. V2 = 0.05 m^3) at constant pressure. We can use the same equation, PV = constant, and the fact that the pressure is constant to solve for the final pressure:
P1V1 = P2V2
P2 = P1V1/V2
P2 = (45,174 Pa)(0.1 m^3)/(0.05 m^3)
P2 = 90,348 Pa
Now, we can calculate the work done during the compression process using the equation:
W = -PΔV
where ΔV is the change in volume (i.e. V2 - V1 = -0.05 m^3), and the negative sign indicates that work is done on the system during compression. Substituting the values, we get:
W = -(45,174 Pa)(-0.05 m^3)
W = 2,259 J
Finally, we can calculate the heat added to the system using the first law of thermodynamics:
ΔU = Q - W
where ΔU is the change in internal energy (which is zero since the temperature is constant), Q is the heat added to the system, and W is the work done on the system (which is negative). Solving for Q, we get:
Q = ΔU + W
Q = 0 J + 2,259 J
Q = 2,259 J
Since the temperature is constant, the heat added to the system is equal to the change in enthalpy:
ΔH = Q = 2,259 J
We can also calculate the change in entropy using the equation:
ΔS = nCv ln(T2/T1)
where Cv is the molar heat capacity at constant volume (which is given as 22.1 J/K.mol), and ln(T2/T1) is the natural logarithm of the ratio of final and initial temperatures. Since the temperature is constant, ΔS = 0.
Therefore, the final answers are:
P1 = 45,174 Pa
P2 = 90,348 Pa
W = 2,259 J
Q = 2,259 J
ΔS = 0