$$PV = nRT$$
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
In this case, we have n = 1.0 mol, V = 2.0 L, R = 0.08206 L·atm/mol·K, and T = 1000 K. Substituting these values into the ideal gas law, we get:
$$P = \frac{nRT}{V} = \frac{(1.0 mol)(0.08206 L·atm/mol·K)(1000 K)}{2.0 L} = 41.03 atm$$
Therefore, the pressure exerted by the gas is 41.03 atm.
