Question

Five moles of an ideal monatomic gas with an initial temperature of 127${}^{\circ }$C expand and, in the process, absorb 1500 J of heat and do 2100 J of work. What is the final temperature of the gas?

Short answer

The final temperature of the gas is 390.5 K.

Step-by-step solution

Convert Initial Temperature to Kelvin

The initial temperature of the gas is given as 127${}^{\circ }$C. To convert this to Kelvin, we add 273.15. Thus, $T_i=127+273.15=400.15\text{K}$.

Apply the First Law of Thermodynamics

The first law of thermodynamics is given by $\mathrm{\Delta }U=Q-W$, where $\mathrm{\Delta }U$ is the change in internal energy, $Q$ is the heat absorbed, and $W$ is the work done by the system. Here, $Q=1500J$ and $W=2100J$. Thus, $\mathrm{\Delta }U=1500-2100=-600J$.

Relate Internal Energy Change to Temperature Change

For an ideal monatomic gas, the change in internal energy can also be expressed using $\mathrm{\Delta }U=\frac{3}{2}nR\mathrm{\Delta }T$, where $n$ is the number of moles and $R$ is the ideal gas constant $8.314J/(mol\cdot K)$.

Solve for Change in Temperature $\mathrm{\Delta }T$

Using $\mathrm{\Delta }U=\frac{3}{2}nR\mathrm{\Delta }T$, we have: $$-600=\frac{3}{2}\times 5\times 8.314\times \mathrm{\Delta }T$$Solve for $\mathrm{\Delta }T$:$$\mathrm{\Delta }T=\frac{-600}{\frac{3}{2}\times 5\times 8.314}$$$$\mathrm{\Delta }T\approx -9.65K$$

Calculate the Final Temperature

Given $T_i=400.15K$ and $\mathrm{\Delta }T=-9.65K$, the final temperature $T_f=T_i+\mathrm{\Delta }T=400.15-9.65=390.5K$.

Convert Final Temperature to Celsius (if needed)

To convert the final temperature from Kelvin to Celsius: $$T_f(\text{degree}C)=390.5-273.15=117.35\text{degree}C$$

Key concepts

first law of thermodynamics

The first law of thermodynamics is a fundamental principle that helps us understand energy changes within a system. It's closely related to the conservation of energy. The law states: $\mathrm{\Delta }U=Q-W$, where:

• $\mathrm{\Delta }U$ is the change in internal energy of the system.
• $Q$ is the heat transferred into the system (positive if absorbed).
• $W$ is the work done by the system on its surroundings (positive if the system does work).

In this exercise, understanding how energy is shared between work and heat is crucial to find the change in internal energy. When the system absorbs 1500 J of heat and does 2100 J of work, it indicates energy is leaving the system, resulting in a negative internal energy change.

internal energy

Internal energy is associated with the microscopic components of a system, like molecules and atoms in a gas. For an ideal monatomic gas, internal energy depends solely on its temperature and is given by: $U=\frac{3}{2}nRT$.
This equation links internal energy directly to the amount of heat added and the work performed.

• $n$ is the number of moles.
• $R$ is the ideal gas constant, $8.314J/(mol\cdot K)$.
• $T$ is the absolute temperature in Kelvin.

As energy is related to the temperature in an ideal monatomic gas, any change in internal energy $\mathrm{\Delta }U$ will result in a temperature change $\mathrm{\Delta }T$, calculated using $\mathrm{\Delta }U=\frac{3}{2}nR\mathrm{\Delta }T$. This relationship helps solve the problem by revealing the decrease in internal temperature based on the energy exchange.

temperature conversion

Converting temperature from Celsius (°C) to Kelvin (K) is often necessary in thermodynamics. This conversion ensures that calculations align with related formulas that require absolute temperatures in Kelvin.
The conversion formula is straightforward: $T(K)=T(°C)+273.15$.
For this exercise:

• The initial temperature is 127°C. Add 273.15 to obtain 400.15 K.
• After calculations, the final temperature of 390.5 K can be converted back to Celsius for practical understanding if needed, using $T(°C)=T(K)-273.15$.
• This results in a final temperature of 117.35°C.

Always remember, Kelvin is essential for scientific calculations, whereas Celsius is often more familiar and suitable for everyday interpretations.

monatomic gas

Monatomic gases consist of individual atoms, like helium or neon, and have properties that simplify thermodynamics calculations. For a monatomic ideal gas, the internal energy and temperature are directly related.

• The equation $U=\frac{3}{2}nRT$ highlights this relationship.
• These gases follow simple behavior due to their lack of molecular bonds, unlike diatomic or polyatomic gases.

In solving thermodynamic problems, knowing the gas's atomic property helps determine the number of degrees of freedom, affecting the calculation of internal energy and, subsequently, temperature changes. This is vital in exercises where temperature and energy change are the focus, like in this example with ideal gas conditions.