Question 659: Thermodynamics & KTG

Question

During the isothermal expansion, a confined ideal gas does $(-150) \mathrm{J}$ of work against its surroundings. This means that

Accepted Answer

Step 1: Recall the First Law of Thermodynamics
$ \Delta U = Q - W $
$ \Delta U $ = change in internal energy
$ Q $ = heat added to the system
$ W $ = work done by the system
Step 2: Isothermal Expansion of an Ideal Gas
For an ideal gas:
$ \Delta U = 0 \quad (	ext{since } \Delta U \propto \Delta T, 	ext{and } \Delta T = 0) $
So:
$ 0 = Q - W \quad \Rightarrow \quad Q = W $
That is, the heat supplied equals the work done by the gas.
Step 3: Interpreting the Signs
The problem states:
The gas does $(-150 \,	ext{J})$ of work.
This implies
using the given sign convention
:
If $W > 0$, work is done
by
the gas.
If $W < 0$, work is done
on
the gas.
So here $W = -150\;	ext{J}$, meaning
the surroundings do 150 J of work on the gas
.
From:
$ Q = W = -150 \,	ext{J} $
So, $Q < 0$. That means
150 J of heat is removed from the gas
.
Ã¢ÂÂ Answer:
Option B. 150 J of heat has been removed from the gas.

Answer

150 J of heat has been added to the gas

Answer

150 J of heat has been removed from the gas

Answer

300 J of heat has been added to the gas

Answer

no heat is transferred because the process is isothermal

Thermodynamics & KTG

During the isothermal expansion, a confined ideal gas does (-150) \mathrm{J} of work against its surroundings. This means that

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