A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 dm 3 to a volume of 20 dm 3. It absorbs 300 J of thermal energy from its surroundings. The Δ U is

Question

A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 dm 3 to a volume of 20 dm 3. It absorbs 300 J of thermal energy from its surroundings. The Δ U is (A) -312 J (B) + 123J (C) -213 J (D) + 231 J

Tardigrade Admin · Accepted Answer

W=-∫ limitsV1V2 P d V=-P(V2-V1) W=-1(20-10)=-10 d m3 atm =-10 d m3 × (8.314 J K-1 m o l-1/0.0821 d m3 K-1 m o l-1) =-1013 J From 1st law of thermodynamic Δ U=q+W =800 J+(-1013 J)=-213 J

Q. A gas expands isothermally against a constant external pressure of $1 a t m$ from a volume of $10 d m^{3}$ to a volume of $20 d m^{3}$ . It absorbs $300 J$ of thermal energy from its surroundings. The $Δ U$ is

-312 J

+ 123J

-213 J

+ 231 J

$W = - V_{1} \int V_{2} P d V = - P (V_{2} - V_{1})$
$W = - 1 (20 - 10) = - 10 d m^{3} a t m$
$= - 10 d m^{3} \times \frac{8.314 J K ^{- 1} m o l ^{- 1}}{0.0821 d m ^{3} K ^{- 1} m o l ^{- 1}}$
$= - 1013 J$
From 1st law of thermodynamic
$Δ U = q + W$
$= 800 J + (- 1013 J) = - 213 J$

Q. A gas expands isothermally against a constant external pressure of 1atm from a volume of 10dm3 to a volume of 20dm3. It absorbs 300J of thermal energy from its surroundings. The ΔU is