A ball with radius 10cm is filled with an ideal gas at pressure 2*(10)^5Pa and temperature 300K. The volume of the gas is changed at constant pressure so that the radius of the ball is reduced with 1cm. Find the amount of gas and the new temperature

In a question like this, you will be given that the molar gas constant is R = 8.31JK^-1mol^-1. However, committing this to memory will be helpful. We should use the ideal gas law: PV = nRT but the volume of the gas is the same as the volume of the ball. The ball has a radius r = 10cm then the volume of the ideal gas in the ball can be found by the formula for a volume of a sphere: V = 4/3πr³ By substitution in the ideal gas law we obtain: P*(4/3πr³) = nRT In order to find the amount of gas in the ball we should find n, thus: n = (4/3*(Pπr³))/(RT) n = (4/3*(2*(10⁵)π(0.1)³)/(8.31×300)= 0.336 mol. In order to find the new temperature T₁ we should use the ideal gas law again but with the new temperature T₁ and volume V₁ of the gas PV₁ = nRT₁ Hence: T₁ = (PV₁)/(nR) but we ought to find the new volume of the ideal gas, which is the same as the new volume of the ball. We can find it by the formula: V₁ =4/3π(r-0.01)³ By substituting the formula for the new volume of the ideal gas in the ideal gas law, we obtain: T₁ = (4/3*(Pπ(r-0.01)³))/(Rn) T₁ = (4/3*(2*(10)⁵π(0.1-0.01)³))/(8.31×0.336)= 218.7K