Question

The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at 1.38 atm (Given, $k = 1.38 \times 10^{-23} \, \text{JK}^{-1}$) is:

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

Answer 1

Answer: (A)

500 K

Answer 2

We use the ideal gas law in terms of the Boltzmann constant:

$PV = NkT$

Where:
- $P = 1.38 \, \text{atm} = 1.38 \times 1.01 \times 10^5 \, \text{Pa}$,
- $N = 2.0 \times 10^{25}$ (total number of molecules),
- $k = 1.38 \times 10^{-23} \, \text{J K}^{-1}$.

Rearranging the formula to solve for $T$:
$T = \frac{PV}{Nk}$

Substituting the values:

$P = 1.38 \times 1.01 \times 10^5 = 1.01 \times 10^5 \, \text{Pa}$

$T = \frac{1.01 \times 10^5}{2 \times 10^{25} \times 1.38 \times 10^{-23}}$

Simplifying, we get:

$T = \frac{1.01 \times 10^3}{2} \approx 500 \, \text{K}$
Thus, the temperature $T$ is 500 K.

The Correct Answer is: 500 K