Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Answer:
a. [tex]P_2=1.38*10^4[mmHg][/tex]
Explanation:
If the Temperature of a gas remains constant (and if the amount of gas molecules doesn't change), then compressing the gas from 12.11 L to 0.669L will increase its pressure through Boyle's Law: [tex]P_1V_1=P_2V_2[/tex]
We'll also need to recall that atmospheric pressure with units of mmHg (since that is the unit requested in the answer) is 760 mmHg.
Letting the initial Pressure and Volume be the P1 and V1, and the final pressure be the P2 and V2, we can substitute and solve:
[tex]P_1V_1=P_2V_2[/tex]
[tex](760[mmHg])(12.11[L])=P_2(0.669[L])[/tex]
[tex]\dfrac{(760[mmHg])(12.11[L \!\!\!\!-])}{0.669[L \!\!\!\!-]}=\dfrac{P_2(0.669[L]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}{0.669[L]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----}}[/tex]
[tex]13757.2496[mmHg]=P_2[/tex]
Since the final pressure is only measured to 3 significant figures, we round the pressure accordingly
[tex]13800[mmHg]=P_2[/tex]
Note that in scientific notation, this is [tex]P_2=1.38*10^4[mmHg][/tex], occasionally written as [tex]1.38\text{E}4[mmHg][/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
