The total average kinetic energy \( E \) of translatory motion for an ideal gas is given by:

\[
E = \frac{3}{2} nRT
\]

where:
- \( n \) is the number of moles,
- \( R \) is the gas constant (\( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.

We can rearrange to find \( nRT \):

\[
nRT = \frac{2}{3} E
\]

The ideal gas law also gives us:

\[
PV = nRT
\]

Thus,

\[
P = \frac{nRT}{V} = \frac{\frac{2}{3} E}{V}
\]

Substitute \( E = 1.5 \times 10^5 \, \text{J} \) and \( V = 20 \, \text{litres} = 20 \times 10^{-3} \, \text{m}^3 \):

\[
P = \frac{\frac{2}{3} \times 1.5 \times 10^5}{20 \times 10^{-3}}
\]

\[
P = \frac{10^5}{20 \times 10^{-3}} = 5 \times 10^6 \, \text{N/m}^2
\]

So, the pressure is \( 5 \times 10^6 \, \text{N/m}^2 \).

For a monoatomic gas under isobaric (constant pressure) conditions, the heat required \( Q \) is given by:

\[
Q = n C_p \Delta T
\]

where:
- \( n = 2 \) moles,
- \( C_p = \frac{5}{2} R \) (specific heat at constant pressure for monoatomic gas),
- \( \Delta T = 50^\circ \text{C} - 0^\circ \text{C} = 50 \) K.

Substitute values:

\[
Q = 2 \times \frac{5}{2} R \times 50 = 250R
\]

So, the answer is \( 250R \).

All the given statements are correct:

• A. Work and heat depend on the path taken during a process, so they are path functions.
• B. Internal energy depends only on the state of the system, making it a state function.
• C. For gases,
  $C_P > C_V$ 

  because extra heat is required at constant pressure to do expansion work.

• D. At constant volume, the gas does no work since volume does not change (
  $W = P\Delta V = 0$ 

  ).