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

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 