# 0.410 mol f a monatomic gas fills a 1 dm^3 container to a pressure of 1.0133 MPa. It’s a expanded reversibly and adiabatically until a pressure of 0.101 33 MPa is reached. What are the final volume and temperature? What is the work done in the expansion?

Please find the solution to the asked query:

The final volume v

_{f}of the gas after adiabatic and reversible expansion can be obtained using the expression:

${p}_{i}{{V}_{i}}^{\gamma}={p}_{f}{{V}_{f}}^{\gamma}\phantom{\rule{0ex}{0ex}}Substitutingthevaluesofpi,{V}_{i},{p}_{f},and\gamma \phantom{\rule{0ex}{0ex}}weget:\phantom{\rule{0ex}{0ex}}(1.0133MPa)(1d{m}^{3}{)}^{5/3}=(0.10133MPa){{V}_{f}}^{5/3}\phantom{\rule{0ex}{0ex}}thusVf={10}^{3/5}d{m}^{3}=3.98d{m}^{3}\phantom{\rule{0ex}{0ex}}ThefinaltemperatureTfaftertheexpansionis:\phantom{\rule{0ex}{0ex}}Tf=\frac{{p}_{f}{V}_{f}}{nR}=\frac{(0.10133\times {10}^{3}kPa)(3.98d{m}^{3})}{(0.410mol)(8.314d{m}^{3}kPamo{l}^{-1}{K}^{-1})}\phantom{\rule{0ex}{0ex}}=118.3K\phantom{\rule{0ex}{0ex}}Theworkdoneduringexpansionis\phantom{\rule{0ex}{0ex}}w=-\frac{{p}_{i}{V}_{i}-{p}_{f}{V}_{f}}{\gamma -1}=-\frac{(1.013MPa)\left(1dm3\right)-(0.10133MPa)(3.98d{m}^{3})}{(5/3)-1}\phantom{\rule{0ex}{0ex}}=-0.915d{m}^{3}MPa=-915J\phantom{\rule{0ex}{0ex}}$

Hope this information will clear your doubts regarding the topic. If you have any other doubt please ask here on the forum and our experts will try to solve them as soon as possible.

Regards

**
**