A gas is allowed to expand in an insulated container against a constant external pressure of $2.5…

A gas is allowed to expand in an insulated container against a constant external pressure of $2.5 \mathrm{~atm}$ from $2.5 \mathrm{~L}$ to $4.5 \mathrm{~L}$, the change in internal energy of the gas in joules is
  1. $-836.3 \mathrm{~J}$
  2. $-1136.2 \mathrm{~J}$
  3. $-450 \mathrm{~J}$
  4. $-506.5 \mathrm{~J}$

Solution

$\mathrm{q}=0$ due to insulated container From first law of thermodynamics, $\begin{aligned} & \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\ & \mathrm{W}=-\operatorname{Pext}\left[\mathrm{V}_2-\mathrm{V}_1\right] \\ & =-2.5[4.5-2.5] \mathrm{atm}-\mathrm{L} \\ & =-2.5 \times 2 \mathrm{~atm}-\mathrm{L} \\ & =-5 \times 101.3 \mathrm{~J} \\ & =-506.5 \mathrm{~J} \\ & \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\ & =0-506.5 \\ & =-506.5 \mathrm{~J} \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

Practice more Chemical Thermodynamics questions on Aicharya