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Sveta_85 [38]
3 years ago
9

The temperature of a sample of gas in a steel tank at 0.30 ATM at 25 Celsius if the pressure W20.59 ATM what would be the final

temperature inside the tank in kelvin
Chemistry
1 answer:
dimulka [17.4K]3 years ago
4 0

The final temperature : 313.06 °C

<h3>Further explanation </h3>

Gay Lussac's Law  

When the volume is not changed, the gas pressure is proportional to its absolute temperature  

\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

P1=0.3 atm

T1=25 + 273=298 K

P2=0.59 atm

\tt \dfrac{0.3}{298}=\dfrac{0.59}{T_2}\\\\T_2=586.06=313.06~^oC

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Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

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Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

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