To determine the temperature of the gas, we can use the Ideal Gas Law in terms of number of molecules. The formula is given by:
\(PV = NkT\)
Where:
The formula can be rearranged to solve for temperature \(T\):
\(T = \frac{PV}{Nk}\)
It is given:
Substitute these values into the rearranged ideal gas formula:
\(T = \frac{1.38 \times 101325}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}\)
Calculate:
\(T = \frac{1.39657 \times 10^{5}}{2.76 \times 10^2}\)
\(T \approx 500 \, \text{K}\)
Thus, the temperature of the gas is 500 K.
The correct answer is: 500 K.
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
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
If \( S \) and \( S' \) are the foci of the ellipse \[ \frac{x^2}{18} + \frac{y^2}{9} = 1 \] and \( P \) is a point on the ellipse, then \[ \min (SP \cdot S'P) + \max (SP \cdot S'P) \] is equal to: