With isobaric heating of 0.1 mol of a monoatomic ideal gas, 4.15 J of heat is consumed.

With isobaric heating of 0.1 mol of a monoatomic ideal gas, 4.15 J of heat is consumed. How many degrees has the gas temperature increased?

Given: ν (amount of substance) = 0.1 mol; Q (heat) = 4.15 J; the process is isobaric.

Constants: i (number of degrees of freedom) = 3; R (gas constant) = 8.31 J / (mol * K).

Let us express the temperature change: Q = ΔU + A = i * (m / M) * R * ΔT / 2 + i * P * ΔV / 2 = i * (m / M) * R * ΔT / 2 + 2 * ΔU / i = i * (m / M) * R * ΔT / 2 + 2 * (i * (m / M) * R * ΔT / 2) / i = i * ν * R * ΔT / 2 + ν * R * ΔT = (i / 2 + 1) * ν * R * ΔT and ΔT = Q / ((i / 2 + 1) * ν * R)

ΔT = 4.15 / ((3/2 + 1) * 0.1 * 8.31) = 2 K.