Question

Find the change in the internal energy of the system in each of situation in Column I.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i. A system absorbs}500\text{cal of heat and}& a.-5000J\\ \text{at the same time does}400J\text{of work}& \\ \text{ii. A system absorbs}300\text{cal and at the}& b.1692J\\ \text{same time}420J\text{of work is done on it}& \\ \text{iii. Twelve hundred calories heat is removed}& c.1680J\\ \text{from a gas held at constant volume}& \end{array}$

• A. i-b, ii-c, iii-a
• B. i-c, ii-b, iii-a
• C. i-b, ii-a, iii-a
• D. i-b, ii-a, iii-c

Answer 1

The correct option is A i-b, ii-c, iii-a

$i\to b.;ii\to c.;iii\to a.$
$i.dQ=dU+dw$
Here
$dQ=+\frac{500\times 4184}{1000}J=2092J$
$dW=+400J$. Since the work is done by the system,
$dU=dQ-dW=2092-400=1692J$
$ii.\Delta U=\Delta Q-\Delta W=\left(300\right)\left(4.184\right)-(-420)=1680J$ (approx).
$iii.\Delta U=(\Delta Q-\Delta W)=(-1200)\left(4.184\right)-0=-5000J$
(Note that $\Delta Q$ is positive when heat is added to the system and $\Delta W$ is positive when the system does the work)

Why this question? Caution: Conversion of calories to joules and vice versa is important and should be remembered.