An ideal gas at 15.5C and a pressure of 1.72 x 10^5 Pa occupies a volume of 2.81 m3. If the volume is raised to 4.16 m3 and the temperature raised to 28.2C, what will be the pressure of the gas?

Explanation

Using the Ideal Gas formula PV=nRT where R is the molar gas constant of 8.31 Joules, with the first set of values (temperature must be converted to Kelvin), the value of n can be found as 200.78. This can then be inputted into the equation with the second set of values (V=4.16 , T=28.2°C=301.35°K) to get Pressure (P) as the number in option B.To solve this problem, we can use the combined gas law:(P1V1)/T1 = (P2V2)/T2where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.Substituting the given values, we get:(1.72 × 10^5 Pa)(2.81 m^3)/(15.5 + 273.15 K) = (P2)(4.16 m^3)/(28.2 + 273.15 K)Solving for P2, we get:P2 = [(1.72 × 10^5 Pa)(2.81 m^3)/(15.5 + 273.15 K)] × [(28.2 + 273.15 K)/(4.16 m^3)] = 1.06 × 10^5 PaTherefore, the pressure of the gas is approximately 1.06 × 10^5 Pa when the volume is raised to 4.16 m^3 and the temperature is raised to 28.2 °C.