Answer:
a. W = 5,708 J, Q = 5,708 J, ΔU = 0 and ΔH = 0.
b. 2,231 J.
Explanation:
Hello!
a. In this case, since this is an isothermal process (constant temperature) it is possible to infer that the work is computed as shown below:
[tex]W=nRTln(\frac{p_1}{p_2} )=1mol*8.314\frac{J}{mol*K}*298.15Kln(\frac{10.bar}{1.0bar} )\\\\W=5708J[/tex]
Now, since this is an isothermal process we know by definition ΔU = 0 and ΔH = 0, therefore the involved heat is:
[tex]Q-W=0\\\\Q=W=5708J[/tex]
b. In this case, since the process is isobaric but goes to the same final volume in part a, we can compute the initial and final volume based on the part a's conditions by using the ideal gas equation:
[tex]V_1=\frac{nRT}{p_1} =\frac{1mol*0.083145\frac{*L}{mol*K}*298.15K}{10.bar} =2.48L\\\\V_2=\frac{nRT}{p_2} =\frac{1mol*0.083145\frac{*L}{mol*K}*298.15K}{1.bar} =24.8L[/tex]
Thus, the work done here is:
[tex]W=p(V_2-V_1)=1bar(24.8L-2.48L)=22.3bar*L*\frac{1x10^5Pa}{1bar} *\frac{1m^3}{1000L} \\\\W=2231J[/tex]
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