Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
