Answer:
Final temperature is 235.4 K
Explanation:
because q=0, ΔU= w
[tex]nC_{v,m} (T_{f} - T_{i}) = -P_{ext} (V_{f} - V_{i})\\[/tex]
substitute for V from ideal gas equation PV = nRT
[tex]nC_{v,m} (T_{f} - T_{i}) = -P_{ext} (\frac{nRT_{f} }{P_{f} } - \frac{nRT_{i}}{P_{i} } )\\\\[/tex]
cancel out factor n and rearrange the equation
[tex]C_{v,m}T_{f} - C_{v,m}T_{i} = -P_{ext}\frac{RT_{f} }{P_{f} } + P_{ext}\frac{RT_{i} }{P_{i} }\\\\C_{v,m}T_{f} + P_{ext}\frac{RT_{f} }{P_{f} } = C_{v,m}T_{i} + P_{ext}\frac{RT_{i} }{P_{i} }\\\\T_{f} (C_{v,m} + \frac{P_{ext}R }{P_{f} } ) = T_{i} (C_{v,m} + \frac{P_{ext}R }{P_{i} } )\\\\[/tex]
make [tex]T_{f}[/tex] the subject of the formula
[tex]T_{f} = T_{i}(\frac{C_{v,m}+\frac{P_{extR} }{P_{i} } }{C_{v,m} + \frac{P_{extR} }{P_{f} } })[/tex]
[tex]T_{i}[/tex] = 20 degrees = 20 + 273K = 293K
R (a constant) = 8.314 Jmol⁻¹K⁻¹
[tex]T_{f} = 293K(\frac{\frac{5}{2}*8.314Jmol^{-1}k^{-1} +\frac{10^{5}Pa * 8.314Jmol^{-1}k^{-1} }{3.20 * 10^{5} Pa } }{\frac{5}{2} * 8.314Jmol^{-1}k^{-1} + \frac{10^{5}Pa * 8.314 Jmol^{-1}k^{-1} }{1*10^{5} Pa } })[/tex]
solve for [tex]T_{f}[/tex] = 235.4 K
The final temperature is 235.4 K
