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 an ideal gas, a cyclic process ABCA as shown in the P–T diagram. When represented in P–V plot, it would be 

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]