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
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.

Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: