What is $\vartriangle \text{U}$ for the presence described by the figure?
Heat supplied during the process, q= 100 KJ.

A. +50 KJ
B. -50 KJ
C. -150 KJ
D. +250 KJ

Answer Verified Verified
Hint: Find the work done using the P-V graph; as the area under the P-V curve represents the work done by the gas. Then use the first law of thermodynamics to calculate $\vartriangle \text{U}$. The first law says that $\vartriangle \text{U}=\text{W+Q}$. Do convert the units into joules.

Complete answer:
Let us first find the work done by the gas using the given graph:
We know that the area under the P-V curve gives the work. This is because the formula of work done is $\text{P}\vartriangle \text{V}$. The area or work in the P-V graph is represented by :

This coloured portion is a trapezium with AB and CD as its parallel sides and height between the two is $\vartriangle \text{V}$. The area of trapezium is sum of parallel sides multiplied to the height and divided by 2 or $\frac{1}{2}\times \left( \text{Sum of parallel sides} \right)\times \left( \text{Height} \right)$. The parallel sides have the value 1 bar and 2 bar respectively. The height is $\vartriangle \text{V}$ or (2-1) ${{\text{m}}^{3}}$ . So, the area will be $\frac{1}{2}\times \left( 1+2 \right)\times \left( 1 \right)$ or 1.5 bar-${{\text{m}}^{3}}$. The work in joules will be 150 KJ as $1\text{ bar-}{{\text{m}}^{3}}=100\text{ KJ}$. Work is positive because $\vartriangle \text{V}$ is positive. So, the value of W is +150 KJ.
- First law of Thermodynamics says that energy can only be converted from one form to another. It can neither be destroyed nor be created. Mathematically, its expression is $\vartriangle \text{U}=\text{W+Q}$.
Q= + 100 KJ, positive because it is added to the system from outside. Using the formula, $\vartriangle \text{U}$ will be $100+150$ or 250 KJ.
The change in internal energy or $\vartriangle \text{U}$ for the process is +250 KJ.

The correct option is ‘d’.

The unit of work done should be known which is joule. The conversion units have to be used correctly. It should be known that if $\vartriangle \text{V}$ is positive, then work will be positive and if $\vartriangle \text{V}$ is negative then, work will also be negative.