Question

The internal energy of a gas decreases by 1.65 kJ when it transfers 1.87 kJ of energy in the form of heat to the surroundings.

(a) Calculate the work done by the gas on the surroundings.
(b) Does the volume of gas increase or decrease?

Answer 1

The first law of thermodynamics states ΔU = q + w, where ΔU is the change in internal energy of a system, q is the heat entering or leaving the system (positive if it is entering the system, negative if it is leaving), and w is work done by the system or on the system (positive if work is being done on the system, negative if the system is doing work on its surroundings).

(a) The problem gives us the change in internal energy, ΔU, -1.65 kJ, as well as change in heat, -1.87 kJ (notice the negative, heat is leaving the system into the surroundings).

Therefore our equation is ΔU = q + p = -1.65 = -1.87 + w.

Solving for w gives us 0.22 kJ.

(b) As stated before, work positive if work is being done on the system and negative if the system is doing work on its surroundings since it represents energy in the form of work in the system. Doing work will transfer that energy out of the system making the sign negative, and having work done on the system transfers that energy into the system, making the sign positive.

Since the sign of 0.22 is positive, that means work is being done on the system by its surroundings. This means the volume of the gas will decrease. If work were negative, the volume would increase.

Answer 2

(a) -0.22 kJ

(b) Decreases

Step-by-step explanation:

To solve this problem, we can use the first law of thermodynamics, which states that the change in the internal energy of a system (ΔE_int.) is equal to the heat added to the system (Q) minus the work done by the system (W_by).The formula is written as:

`\Delta E_{\text{int.}}=Q-W_{\text{by}}`

We are given:

• ΔE_int. = -1.65 kJ ("-" since energy decreases)
• Q = -1.87 kJ ("-" since heat is leaving the system)

(a) To find the work done by the gas on the surroundings, we rearrange the formula to solve for 'W_by':

`\Longrightarrow W_{\text{by}}=Q-\Delta E_{\text{int.}}`

Plug in our values and solve:


`\Longrightarrow W_{\text{by}}=-1.87 \text{ kJ}-(-1.65 \text{ kJ})\\\\\\\\\Longrightarrow W_{\text{by}}=-1.87 \text{ kJ}+1.65 \text{ kJ}\\\\\\\\\therefore W_{\text{by}}=\boxed{-0.22 \text{ kJ}}`

Thus, the work done by the gas on the surroundings is -0.22 kJ. The negative sign indicates that the surroundings do work on the gas.

(b) Since the work done by the gas is negative, it means that the surroundings have done work on the gas. This typically occurs when the volume of the gas decreases.

We can prove this using the following formula for calulating the work done by a gas:

`W_{\text{by}}=P\Delta V`

Where:

• 'W_by' is the work done by the gas
• 'P' is the pressure
• 'ΔV' is the change in volume

If the pressure is held constant we can use sign convention. If 'W_by' was positive this would indicate an increase in volume. Since 'W_by' was negative, this indicated that there was a decrease in volume.