Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A gas sample occupies \mathrm{ 2.5 \mathrm{~L}} at \mathrm{ 27^{\circ} \mathrm{C}} and 1 bar pressure. If the gas undergoes an isothermal expansion to \mathrm{ 10 \mathrm{~L}}, what is the final pressure?

Option: 1

0.25 bar


Option: 2

0.42 bar


Option: 3

0.6 bar


Option: 4

0.8 bar


Answers (1)

best_answer

Step 1: Use the ideal gas law to calculate the initial number of moles \mathrm{\left(n_1\right)} of the gas.

\mathrm{ \begin{gathered} P V=n R T \\\\ n_1=\frac{1 \mathrm{bar} \times 2.5 \mathrm{~L}}{0.0831 \mathrm{~L} \cdot \mathrm{bar} / \mathrm{mol} \cdot \mathrm{K} \times 300 \mathrm{~K}} \approx 0.1004 \mathrm{~mol} \end{gathered} }

Step 2: For isothermal condition we can use the final volume \mathrm{\left(V_2\right)} and the initial number of moles \mathrm{\left(n_1\right)} to calculate the final pressure \mathrm{\left(P_2\right).}

\mathrm{ \begin{gathered} P_1 V_1=n_1 R T \\\\ P_2=\frac{n_1 R T}{V_2} \\\\ P_2=\frac{0.1004 \mathrm{~mol} \times 0.0831 \mathrm{~L} \cdot \mathrm{bar} / \mathrm{mol} \cdot \mathrm{K} \times 300 \mathrm{~K}}{10 \mathrm{~L}} \approx 0.2497 \mathrm{bar} \end{gathered} }

Therefore, the correct option is A.

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

75% criteria for JEE Main 2025 not removed; eligibility criteria for admission to IITs
anam-khan

BTech Admissions 2025 LIVE: AEEE phase 1 exam date out, JEE Mains registration ongoing; BITSAT, VITEEE updates
anam-khan

JEE Main 2025 Eligibility Criteria: What is state code of eligibility? Here’s list of qualifying exams

Latest Question