1 mol of an ideal monoatomic gas undergoes isothermal expansion from volume $V_1$ to $V_2 = 2V_1$ at temperature $T = 300$ K. Find the work done.
1 mol of an ideal monoatomic gas undergoes isothermal expansion from volume $V_1$ to $V_2 = 2V_1$ at temperature $T = 300$ K. Find the work done.
1 Answer
NANisha Agarwal
✓ Accepted
· 2mo ago
▲ 28
For isothermal process:
$$W = nRT\ln\frac{V_2}{V_1} = 1 \times 8.314 \times 300 \times \ln 2$$
$$W = 2494.2 \times 0.693 = 1728 J$$
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Discussion (3)
PI
Clean and to the point. Bookmarking this for revision.
O
This is exactly the kind of step-by-step I needed. Respect.
J
I solved it a slightly different way and got the same answer, good sign.