The temperature of a gas having `2.0 times 10^(25)` molecules per cubic meter at 1.38 atm (Given, `k = 1.38 times 10^(–23) JK^(–1)`) is :

Options

A. 500 K
B. 200 K
C. 100 K
D. 300 K

Answer

To find the temperature of the gas, we can use the ideal gas law in the form of the equation that relates pressure, density, and temperature. The relevant equation is:

\[ P = \frac{n}{V} k T \]

Where:
- \( P \) is the pressure,
- \( n/V \) is the number density of molecules (number of molecules per unit volume),
- \( k \) is the Boltzmann constant,
- \( T \) is the temperature in Kelvin.

Given:
- Pressure \( P = 1.38 \, \text{atm} \)
- Number density \( n/V = 2.0 \times 10^{25} \, \text{m}^{-3} \)
- Boltzmann constant \( k = 1.38 \times 10^{-23} \, \text{JK}^{-1} \)

### Step 1: Convert Pressure to Pascals
First, we need to convert the pressure from atm to Pascals. The conversion factor is:
\[ 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \]

So,
\[ P = 1.38 \, \text{atm} \times 1.013 \times 10^5 \, \text{Pa/atm} = 1.396 \times 10^5 \, \text{Pa} \]

### Step 2: Rearrange the Ideal Gas Equation
Rearranging the equation to solve for temperature \( T \):
\[ T = \frac{P}{(n/V) k} \]

### Step 3: Substitute the Values
Now we substitute the values into the equation:
\[ T = \frac{1.396 \times 10^5 \, \text{Pa}}{(2.0 \times 10^{25} \, \text{m}^{-3})(1.38 \times 10^{-23} \, \text{JK}^{-1})} \]

### Step 4: Calculate the Denominator
Calculating the denominator:
\[ (2.0 \times 10^{25}) \times (1.38 \times 10^{-23}) = 2.76 \times 10^{2} \]

### Step 5: Calculate Temperature
Now substituting back into the equation for \( T \):
\[ T = \frac{1.396 \times 10^5}{2.76 \times 10^2} \]

Calculating this gives:
\[ T \approx 505.8 \, \text{K} \]

### Step 6: Round to Appropriate Significant Figures
Rounding to two significant figures, we get:
\[ T \approx 500 \, \text{K} \]

### Final Answer
The temperature of the gas is approximately **500 K**.

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