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Q.
$ 1 $ mole of gas expands isothermally at $ 37^{\circ}C $ . The amount of heat is absorbed by it until its volume doubled is $ (R = 8.31 \,J \,mol^{-1} K^{-1}) $
Thermodynamics
Solution:
Here, $ T= 37 ^{\circ} C = 37 + 273 = 310\, K, V_2 = 2V_1 $
work done in isothermal process is
$ W =nRT ln \frac{V_2}{V_1} $
$ = 1\times 8.31 \times 310 \times ln \frac{2V_1}{V_1} $
$ =8.31 \times 310 \times ln 2 = 1.786 \times 10^3 $
Amount of heat absorbed $ = \frac{1.786 \times 10^3}{4.2} $ cal
$ = 425.4 $ cal