A constant - volume gas thermometer is used to measure the temperature of an object . When the thermometer is in contact with water at its triple point (273K) the pressure in the thermometer is `8.50 xx10^(4) Pa` . When it is contact with the object the pressure is `9.650 xx10^(4) Pa` . The real temperature of the object is

Options

A. 41.0 K
B. 114 K
C. 241 K
D. 310 K

Answer

To find the real temperature of the object using the constant-volume gas thermometer, we can use the relationship between pressure and temperature given by the formula:

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Where:
- \( P_1 \) is the pressure at the triple point of water,
- \( T_1 \) is the temperature at the triple point of water,
- \( P_2 \) is the pressure when in contact with the object,
- \( T_2 \) is the temperature of the object we want to find.

### Step 1: Identify the known values
- \( P_1 = 8.50 \times 10^4 \, \text{Pa} \)
- \( T_1 = 273 \, \text{K} \)
- \( P_2 = 9.650 \times 10^4 \, \text{Pa} \)

### Step 2: Rearrange the formula to solve for \( T_2 \)
From the formula, we can rearrange it to find \( T_2 \):

\[ T_2 = \frac{P_2 \cdot T_1}{P_1} \]

### Step 3: Substitute the known values into the equation
Now, substitute the known values into the rearranged formula:

\[ T_2 = \frac{(9.650 \times 10^4 \, \text{Pa}) \cdot (273 \, \text{K})}{8.50 \times 10^4 \, \text{Pa}} \]

### Step 4: Perform the calculations
First, calculate the numerator:

\[ 9.650 \times 10^4 \times 273 = 2.63445 \times 10^7 \]

Now, divide by \( P_1 \):

\[ T_2 = \frac{2.63445 \times 10^7}{8.50 \times 10^4} \approx 310 \, \text{K} \]

### Final Answer
The real temperature of the object is approximately \( T_2 = 310 \, \text{K} \).

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