To solve the problem of determining the temperature needed to double the pressure of a gas within a fixed volume, we apply the Ideal Gas Law in the form of Charles's law, which is given by:
\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
Here, \(P_1\) is the initial pressure, \(T_1\) is the initial temperature in Kelvin, \(P_2\) is the final pressure, and \(T_2\) is the final temperature in Kelvin.
The initial condition is at 27°C, which is 300K (since \(T(K) = T(°C) + 273\)). We aim to double the pressure (\(P_2 = 2P_1\)).
Solving for \(T_2\):
\[ \frac{P_1}{300} = \frac{2P_1}{T_2} \]
By simplifying, we find:
\[ T_2 = 2 \times 300 = 600 \, \text{K} \]
Convert back to Celsius: \(T(°C) = T(K) - 273\):
\(T_2 = 600 - 273 = 327°C\).
Thus, the temperature should be raised to 327°C to double the pressure.

The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.