Question

To what temperature must 15 L of oxygen gas at 32oC be heated at 1 atm pressure in order to occupy a volume of 23 L, assuming that the pressure increases by 31 mm Hg?

Answer 1

To find the temperature at which the oxygen gas must be heated, we can use the Ideal Gas Law, PV = nRT. Given the initial conditions of pressure, volume, and temperature, we can calculate the final temperature by rearranging the equation and substituting the values. In this case, the final temperature is approximately 514.56 K.

Step-by-step explanation:

To find the temperature to which the oxygen gas must be heated, we can use the Ideal Gas Law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Given the initial conditions: P1 = 1 atm, V1 = 15 L, and T1 = 32°C, we need to convert the temperature to Kelvin by adding 273.15:

T1 = 32 + 273.15 = 305.15 K.

Since the question states that the pressure increases by 31 mm Hg, we can convert this to atm by dividing by 760: ΔP = 31 mm Hg / 760 atm/mm Hg

= 0.0408 atm.

Using the equation PV = nRT,

we can rearrange it to solve for T2 (the final temperature) and substitute the given values:

T2 = (P2 * V1 * T1) / (P1 * V2)

= (1.0408 atm * 15 L * 305.15 K) / (1 atm * 23 L)

= 514.56 K.