Answer:
29273.178 joules have been added to the gas for the entire process.
Explanation:
The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:
Isochoric (Constant volume)
[tex]c_{v} = \frac{3}{2}\cdot R_{u}[/tex] (1)
Isobaric (Constant pressure)
[tex]c_{p} = \frac{5}{2}\cdot R_{u}[/tex] (2)
Where [tex]R_{u}[/tex] is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.
Under the assumption of ideal gas, we notice the following relationships:
1) Temperature is directly proportional to pressure.
2) Temperature is directly proportional to volume.
Now we proceed to find all required temperatures below:
(i) Alice heats the gas at constant volume until its pressure is doubled:
[tex]\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}[/tex] (3)
([tex]\frac{P_{2}}{P_{1}} = 2[/tex], [tex]T_{1} = 273.15\,K[/tex])
[tex]T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}[/tex]
[tex]T_{2} = 2\times 273.15\,K[/tex]
[tex]T_{2} = 546.3\,K[/tex]
(ii) Bob further heats the gas at constant pressure until its volume is doubled:
[tex]\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}[/tex] (4)
([tex]\frac{V_{3}}{V_{2}} = 2[/tex], [tex]T_{2} = 546.3\,K[/tex])
[tex]T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}[/tex]
[tex]T_{3} = 2\times 546.3\,K[/tex]
[tex]T_{3} = 1092.6\,K[/tex]
Finally, the heat added to the gas ([tex]Q[/tex]), measured in joules, for the entire process is:
[tex]Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})][/tex] (5)
If we know that [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], [tex]n = 2\,mol[/tex], [tex]T_{1} = 273.15\,K[/tex], [tex]T_{2} = 546.3\,K[/tex] and [tex]T_{3} = 1092.6\,K[/tex], the heat added to the gas for the entire process is:
[tex]c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)[/tex]
[tex]c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]
[tex]c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)[/tex]
[tex]c_{p} = 20.785\,\frac{J}{mol\cdot K}[/tex]
[tex]Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right][/tex]
[tex]Q = 29273.178\,J[/tex]
29273.178 joules have been added to the gas for the entire process.
