Answer:
**Problem 1:**
The temperature of a substance is 25°C. What is the temperature in Kelvin (K)?
**Solution 1:**
To convert Celsius to Kelvin, use the formula:
\[ K = °C + 273.15 \]
Given \( °C = 25 \):
\[ K = 25 + 273.15 = 298.15 \, K \]
**Answer 1:**
The temperature is 298.15 Kelvin.
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**Problem 2:**
A gas is initially at a temperature of 300 K. If its temperature increases by 50°C, what is the final temperature in Kelvin?
**Solution 2:**
Convert Celsius to Kelvin and then add to the initial Kelvin temperature.
\[ K_f = K_i + \Delta °C \]
\[ K_f = 300 + (50 + 273.15) \]
\[ K_f = 300 + 323.15 \]
\[ K_f = 623.
**Problem 3:**
A sample of gas occupies a volume of 2.50 liters at a temperature of 27°C. If the pressure remains constant, what will be the volume of the gas in Kelvin if the temperature is increased to 300 K?
**Solution 3:**
Convert Celsius to Kelvin for the initial temperature:
\[ T_{initial} = 27°C + 273.15 = 300.15 K \]
Now, using the combined gas law \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \):
\[ \frac{P \cdot 2.50}{300.15} = \frac{P \cdot V_2}{300} \]
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**Problem 4:**
A metal rod has a length of 50 cm at 20°C. If the rod expands uniformly, what will be its length in Kelvin if the temperature increases to 100°C? Assume a linear expansion coefficient of \(1.2 \times 10^{-5} \) per degree Celsius.
**Solution 4:**
First, convert the temperatures to Kelvin:
- Initial temperature, \(T_1 = 20°C + 273.15 = 293.15 K\)
- Final temperature, \(T_2 = 100°C + 273.15 = 373.15 K\)
Use the formula for linear thermal expansion:
\[ \Delta L = L_0 \alpha (T_2 - T_1) \]
Where:
- \( L_0 \) is the initial length of the rod (50 cm)
- \( \alpha \) is the linear expansion coefficient ( \(1.2 \times 10^{-5} \) per °C)
Calculate \( \Delta L \):
\[ \Delta L = 50 \times 10^{-2} \times (1.2 \times 10^{-5}) \times (373.15 - 293.15) \]
\[ \Delta L = 50 \times 10^{-2} \times 1.2 \times 10^{-5} \times 80 \]
\[ \Delta L = 0.048 \times 10^{-2} \]
\[ \Delta L = 0.48 \text{ cm} \]
Now, find the final length of the rod:
\[ L_{final} = L_0 + \Delta L \]
\[ L_{final} = 50 + 0.48 \]
\[ L_{final} = 50.48 \text{ cm} \]
**Answer 4:**
The length of the rod at 100°C (373.15 K) is 50.48 cm.
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**Problem 5:**
A certain gas occupies a volume of 2.00 liters at 27°C and 1 atm pressure. If the volume remains constant, what will be the temperature in Kelvin if the pressure increases to 2 atm?
**Solution 5:**
Convert Celsius to Kelvin for the initial temperature:
\[ T_{initial} = 27°C + 273.15 = 300.15 K \]
Apply the ideal gas law, \( P_1 \cdot V_1 / T_1 = P_2 \cdot V_2 / T_2 \), where V
