Calculate Q and w for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K .

Question

Calculate Q and w for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K . (A) 5.22 kJ,-5.22 kJ (B) -5.22 kJ,5.22 kJ (C) 27.3 kJ,-27.3 kJ (D) -27.3 kJ,27.3 kJ

Tardigrade Admin · Accepted Answer

Δ w=2.303× nRTlog(P1/P2) Δ U=0 -q=w wr e v=-2.303× 1× 8.314× 273log(1/0.1)=-5227 J wr e v .=-5.22 kJ

Q. Calculate $Q$ and $w$ for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of $1.0$ bar to a final pressure of $0.1$ bar at a constant temperature of $273$ $K$ .

$5.22$ $k J, - 5.22$ $k J$

$- 5.22$ $k J, 5.22$ $k J$

$27.3$ $k J, - 27.3$ $k J$

$- 27.3$ $k J, 27.3$ $k J$

$Δ w = 2.303 \times n RTl o g \frac{P _{1}}{P _{2}}$

$Δ U = 0$

$- q = w$

$w_{re v} = - 2.303 \times 1 \times 8.314 \times 273 l o g \frac{1}{0.1} = - 5227$ $J$

$w_{re v .} = - 5.22$ $k J$

Q. Calculate Q and w for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K .

5.22 kJ,−5.22 kJ

−5.22 kJ,5.22 kJ

27.3 kJ,−27.3 kJ

−27.3 kJ,27.3 kJ