Question

A poorly engineered system uses He (heat capacity of 12.5 J K^-1 mol^-1) as a coolant for a highly exothermic reaction which releases 23.1 kJ of heat. If the volume of the container holding 10 moles of He is fixed, how much does the pressure change in the coolant tube when the He absorbs the released heat? Assume all the heat released by the reaction is absorbed by the He and its initial temperature is 10.0 °C.

Answer 1

The formula for heat absorbed by the coolant is q = nc∆T where n is the number of moles of coolant, c is the heat capacity, ∆T is the change in temperature.

23100 J = 10 mol* 12.5 J/K.mol * ∆T

∆T = 184.8 K

∆T = Tf - Ti

Ti = 10°C = 283 K

Tf = 184.8 + 283 = 467.8 K

Let the initial pressure be 1 atm. When volume is constant, P is directly proportional to T.

P1/P2 = T1/T2

1 atm / P2 = 283 K /467.8 K

P2 = 467.8 / 283 = 1.653 atm

Change in pressure = 1.653 - 1 = 0.653 atm

Percentage change = (0.653 atm/1)*100 = 65.3%

The pressure increases by 65.3%