1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of 15 L. The work done by a system is equal to

Question

1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of 15 L. The work done by a system is equal to (A) -1.215 × 103 J (B) -12.15 × 103 J (C) +1.215 × 103 J (D) +12.15 × 103 J

Tardigrade Admin · Accepted Answer

p ext =1 atm V1=3 L ⇒ V2=3 L As work is done against constant external pressure, the process is irreversible. W =-p ext Δ V =-1 atm [15-3] L=-12 atm L =-12 × 101.3 J [11 atm =101.3 J ] =-1215.6 J =-1.2156 × 103 J

Q. $1$ mole of gas occupying $3 L$ volume is expanded against a constant external pressure of $1$ atm to a volume of $15 L$ . The work done by a system is equal to

$- 1.215 \times 1 0^{3} J$

$- 12.15 \times 1 0^{3} J$

$+ 1.215 \times 1 0^{3} J$

$+ 12.15 \times 1 0^{3} J$

$p_{e x t} = 1 a t m$

$V_{1} = 3 L$

$\Rightarrow V_{2} = 3 L$

As work is done against constant external pressure, the process is irreversible.

$W = - p_{e x t} Δ V$

$= - 1 a t m [15 - 3] L = - 12 a t m L$

$= - 12 \times 101.3 J [11 a t m = 101.3 J]$

$= - 1215.6 J = - 1.2156 \times 1 0^{3} J$

Q. 1 mole of gas occupying 3L volume is expanded against a constant external pressure of 1 atm to a volume of 15L. The work done by a system is equal to

−1.215×103J

−12.15×103J

+1.215×103J

+12.15×103J

pext​=1atm V1​=3L ⇒V2​=3L As work is done against constant external pressure, the process is irreversible. W=−pext​ΔV =−1atm[15−3]L=−12atmL =−12×101.3J[11atm=101.3J] =−1215.6J=−1.2156×103J

Q. $1$ mole of gas occupying $3 L$ volume is expanded against a constant external pressure of $1$ atm to a volume of $15 L$ . The work done by a system is equal to

$- 1.215 \times 1 0^{3} J$

$- 12.15 \times 1 0^{3} J$

$+ 1.215 \times 1 0^{3} J$

$+ 12.15 \times 1 0^{3} J$