Question

A system absorbs 100 kJ heat in the
process shown in the figure. What is∆U
for the system?
a) _50kj
b)+50kj
c)+150kj
d)-150kj​

Answer 1

Answer:

Explanation:

work done = area under the graph

                   = 1/2 * 3 * 1

                  = 1.5 bar m3 = 1.5 x 101 = 150 joules

dq =d u +d w

du = 100 - 150

   = -50 joules

For physics and chem he equation changes that you know isn't it ?

Answer 2

Answer:

The change in the internal energy, ΔU, for the system is $-50kJ$.

Therefore, option a) $-50kJ$ is the correct choice.

Explanation:

Given,

The heat absorbed by the system, Q = $100kJ$

The change in internal energy, ΔU =?

As we know,

• From the first law of thermodynamics, the change in internal energy  has given by the equation:
• $\Delta U = Q +W$     -------equation (1)

Here,

• ΔU = Change in the internal energy
• Q = Heat absorbed by the system ( positive )
• W = Work done on the system ( negative )

Now, firstly, we have to calculate the work done on the system by using the area under the graph:

• Work done = $- [(\frac{1}{2}\times base\times height) + ( area of rectangle)]$ ------equation(2)

As shown in the graph:

• Base = $1m^{3}$
• Height = $1 bar$
• Area of rectangle = $length \times breadth$ = $1 bar\times 1m^{3}$ = $1 bar-m^{3}$.

After putting the values in equation (2), we get:

• Work done = $- [(\frac{1}{2}\times 1\times 1) + (1)]$
• Work done = $-\frac{3}{2} bar-m^{3}$ = $-1.5 bar-m^{3}$

Convert bar-m³ into kJ

• 1  bar-m³ = 100 kJ
• $-1.5 bar-m^{3}$ = $(-1.5 \times 100) kJ$ = $-150kJ$

Now, after putting the value of work done in the equation (1), we get:

• $\Delta U = Q +W$
• $\Delta U = 100 + (- 150)$
• $\Delta U = -50 kJ$

Hence, the change in internal energy, ΔU = $-50kJ$