1. Tardigrade
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  4. A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208 J of heat. The values of q and W for the process will be (R = 8.314 J /mol K) ( ln 7.5 = 2.01)

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Q. A piston filled with $0.04$ mole of an ideal gas expands reversibly from $50.0\, mL$ to $375 \,mL$ at a constant temperature of $37.0^{\circ}C$. As it does so, it absorbs $208\, J$ of heat. The values of $q$ and $W$ for the process will be
$(R = 8.314\,J /mol\, K) (\ln 7.5 = 2.01)$

JEE MainJEE Main 2013Thermodynamics

Solution:

The process is isothermal expansion, hence
$q =-W $
$\Delta E =0$
$W =-2.303 n R T \log \frac{V_{2}}{V_{1}}$
$=-2.303 \times 0.04 \times 8.314 \times 310 \times \log \frac{335}{50} $
$=-208\, J $
$ q =+208\, J , $
$ W=-208 \,J $ (expansion Work )