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)

Question

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) (A) q = + 208 J,W = - 208 J (B) q = - 208 J, W = - 208 J (C) q = - 208 J, W = + 208 J (D) q = + 208 J, W = + 208 J

Tardigrade Admin · Accepted Answer

The process is isothermal expansion, hence q =-W Δ E =0 W =-2.303 n R T log (V2/V1) =-2.303 × 0.04 × 8.314 × 310 × log (335/50) =-208 J q =+208 J , W=-208 J (expansion Work )

Q. A piston filled with $0.04$ mole of an ideal gas expands reversibly from $50.0 m L$ to $375 m L$ 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
$(R = 8.314 J / m o l K) (ln 7.5 = 2.01)$

$q = + 208 J, W = - 208 J$

$q = - 208 J, W = - 208 J$

$q = - 208 J, W = + 208 J$

$q = + 208 J, W = + 208 J$

The process is isothermal expansion, hence
$q = - W$
$Δ E = 0$
$W = - 2.303 n RT lo g \frac{V _{2}}{V _{1}}$
$= - 2.303 \times 0.04 \times 8.314 \times 310 \times lo g \frac{335}{50}$
$= - 208 J$
$q = + 208 J,$
$W = - 208 J$ (expansion Work )

Q. A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0mL to 375mL at a constant temperature of 37.0∘C. As it does so, it absorbs 208J of heat. The values of q and W for the process will be (R=8.314J/molK)(ln7.5=2.01)

q=+208J,W=−208J

q=−208J,W=−208J

q=−208J,W=+208J

q=+208J,W=+208J