The final temperature of the gas is k
When studying thermodynamics, one of the most common questions students encounter is: what is the final temperature of the gas after a particular process? Whether the gas is heated in a piston, expanded adiabatically, or cooled at constant volume, determining the final temperature in kelvin (K) is essential for predicting pressure, volume, and energy changes. This article walks through the concepts, equations, and practical steps needed to calculate the final temperature of a gas, providing clear examples and troubleshooting tips that will help you master the topic for exams, laboratory work, or real‑world engineering applications.
Not the most exciting part, but easily the most useful.
Understanding Gas Temperature and the Kelvin Scale
Temperature measures the average kinetic energy of the particles in a substance. But for gases, the kelvin (K) scale is the absolute unit used in scientific equations because it starts at absolute zero—the point where molecular motion theoretically ceases. Unlike Celsius or Fahrenheit, kelvin values are always non‑negative, which simplifies the ideal gas law and related thermodynamic expressions.
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Key points to remember
- 0 K = –273.15 °C (absolute zero).
- The size of one kelvin is identical to one degree Celsius; only the offset differs.
- In equations such as (PV = nRT), T must be expressed in kelvin; using Celsius will give incorrect results.
When a problem states “the final temperature of the gas is k,” it implicitly asks you to solve for T_f (final temperature) and report the answer in kelvin.
The Ideal Gas Law and Temperature Change
The ideal gas law provides the simplest relationship among pressure (P), volume (V), amount of substance (n), the universal gas constant (R), and temperature (T):
[ PV = nRT ]
If the amount of gas and the gas constant remain unchanged, any change in pressure or volume directly affects temperature. Rearranging the equation gives:
[ T = \frac{PV}{nR} ]
Thus, to find the final temperature after a process, you need the final pressure (P_f) and final volume (V_f), along with the known number of moles (n) and R = 8.314 J mol⁻¹ K⁻¹.
When to Use the Combined Gas Law
If the number of moles does not change, the ideal gas law can be reduced to the combined gas law:
[ \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f} ]
This form is especially handy because it eliminates the need to know n or R explicitly; you only need the initial state (P_i, V_i, T_i) and two of the three final variables to solve for the third Most people skip this — try not to..
Common Thermodynamic Processes and Their Temperature Formulas
Different processes impose specific constraints on P, V, or T, leading to simplified formulas for the final temperature. Below are the most frequently encountered cases.
1. Isochoric (Constant Volume) Process
- Constraint: (V_f = V_i)
- Formula: (\displaystyle \frac{P_f}{P_i} = \frac{T_f}{T_i}) → (T_f = T_i \frac{P_f}{P_i})
If the gas is heated in a rigid container, the pressure rise directly mirrors the temperature rise Not complicated — just consistent..
2. Isobaric (Constant Pressure) Process
- Constraint: (P_f = P_i)
- Formula: (\displaystyle \frac{V_f}{V_i} = \frac{T_f}{T_i}) → (T_f = T_i \frac{V_f}{V_i})
Expanding a gas at constant pressure (e.g., a piston with a fixed weight) increases its temperature proportionally to the volume increase.
3. Isothermal (Constant Temperature) Process
- Constraint: (T_f = T_i)
- Note: Temperature does not change; any change in P or V is compensated by the other.
4. Adiabatic (No Heat Exchange) Process
- Constraint: (Q = 0)
- Relation: (PV^{\gamma} = \text{constant}), where (\gamma = C_p/C_v) (heat‑capacity ratio).
- Temperature‑Volume relation: (\displaystyle T_f V_f^{\gamma-1} = T_i V_i^{\gamma-1}) → (T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma-1})
- Temperature‑Pressure relation: (\displaystyle T_f = T_i \left(\frac{P_f}{P_i}\right)^{(\gamma-1)/\gamma})
Adiabatic expansions cool the gas, while compressions heat it, because work is done at the expense of internal energy.
5. Polytropic Process
- Constraint: (PV^n = \text{constant}) (n is the polytropic index).
- Temperature relation: (\displaystyle T_f = T_i \left(\frac{V_i}{V_f}\right)^{n-1} = T_i \left(\frac{P_f}{P_i}\right)^{(n-1)/n})
When n = 1, the process is isothermal; when n = γ, it is adiabatic Took long enough..
Step‑by‑Step Calculation Examples
Below are three worked‑out problems that illustrate how to determine the final temperature of a gas in kelvin for different scenarios. Follow the outlined steps, and you’ll be able to tackle similar questions on your own.
Example 1: Isochoric Heating
Problem: A 2.0 mol sample of nitrogen gas is confined in a rigid 10 L container at an initial pressure of 1.0 atm and temperature of 300 K. The gas is heated until the pressure reaches 2.5 atm. What is the final temperature?
Solution:
- Identify the process: constant volume → isochoric.
- Use the isochoric relation: (T_f = T_i \frac{P_f}{P_i}).
- Plug in values: (T_f = 300,\text{K} \times \frac{2.5,\text{atm}}{1.0,\text{atm}} = 300 \times 2.5 = 750,\text{K}).
Answer: The final temperature is 750 K.
Example 2: Isobaric Expansion
Problem: Carbon dioxide (0.5 mol) at 400 K occupies a volume of 5.0 L under a constant pressure of 2.0 atm. The gas expands to a volume of 12.0 L while the pressure remains unchanged. Find the final temperature.
Solution:
Solution:
- Identify the process: The pressure is held constant, so the transformation is isobaric.
- Select the governing relation: For an isobaric change of an ideal gas, temperature scales directly with volume:
[ \frac{V_f}{V_i}= \frac{T_f}{T_i};;\Longrightarrow;;T_f = T_i\frac{V_f}{V_i} ]
- Insert the numerical data:
[ T_i = 400;\text{K},\qquad V_i = 5.0;\text{L},\qquad V_f = 12.0;\text{L} ]
[ T_f = 400;\text{K}\times\frac{12.0;\text{L}}{5.0;\text{L}} = 400;\text{K}\times 2.
- (Optional) Verify with the ideal‑gas law: Because the pressure stays at 2.0 atm, the product (nRT) must increase in the same proportion as the volume. Using (n=0.5) mol and (R=0.08206;\text{L·atm·K}^{-1}\text{mol}^{-1}):
[ P V_i = nR T_i ;;\Rightarrow;; 2.0;\text{atm}\times5.0;\text{L}=0.5;\text{mol}\times0.08206;\frac{\text{L·atm}}{\text{K·mol}}\times400;\text{K} ]
[ P V_f = nR T_f ;;\Rightarrow;; 2.0;\text{atm}\times12.0;\text{L}=0.5;\text{mol}\times0.08206;\frac{\text{L·atm}}{\text{K·mol}}\times960;\text{K} ]
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