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Problem 10.3 A sample of N,was placed in a flexible 9.0 L containerat 300K at a pressure of 1.5 am Thegas was compressed to a volume of3.0L and heat was added until thetemperature reached 600K. What is thenew pressure inside the container?Given Data |
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Answer» The new pressure inside the container is 9 atm Explanation: 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} T 1
P 1
V 1
= T
P 2
V 2
where, P_1P 1
= initial pressure of gas = 1.5 atm P_2P 2
= final pressure of gas = ? V_1V 1
= initial volume of gas = 9.0 L V_2V 2
= final volume of gas = 3.0 L T_1T 1
= initial temperature of gas = 300 K T_2T 2
= final temperature of gas = 600 K Now put all the given values in the above equation, we get: \frac{1.5atm\times 9.0L}{300K}=\frac{P_2\times 3.0L}{600K} 300K 1.5atm×9.0L
= 600K P 2
×3.0L
P_2=9atmP 2
=9ATM |
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