A volume of 140 cm3 of an ideal gas has an initial temperature of 20°C and an initial pressure of 1 atm. What is the final pressure if the volume is reduced to 120 cm3 and the temperature is raised to 40°C?

PV = kT, so PV/T is constant.

So, you want P such that

(P)(120)/(40+273) = (1)(140)/(20+273)

To solve this problem, we can use the ideal gas law, which states that the product of the pressure (P), volume (V), and temperature (T) of a gas is constant, expressed as PV = nRT, where n is the number of moles of the gas and R is the ideal gas constant.

Since we are given the pressure, volume, and temperature before and after the change, we can set up two equations and solve for the final pressure (Pf). Let's use subscripts '1' and '2' to represent the initial and final states, respectively.

Equation 1:
P1 * V1 / T1 = P2 * V2 / T2

Plugging in the given values:
1 atm * 140 cm^3 / (20 + 273.15 Kelvin) = Pf * 120 cm^3 / (40 + 273.15 Kelvin)

Simplifying the equation:
140 / (20 + 273.15) = Pf * 120 / (40 + 273.15)

Now, let's solve for Pf:
Pf = (140 / (20 + 273.15)) * (40 + 273.15) * (120 / 140)

Using a calculator, we can evaluate the expression:
Pf ≈ 1.584 atm

Therefore, the final pressure, when the volume is reduced to 120 cm3 and the temperature is raised to 40°C, is approximately 1.584 atm.