Question

A container of nitrogen gas at 23 degrees celcius contains 425 L at a pressure of 3.5 atm. If 26.6 kJ of heat are added to the container, what will be the new temperature of the gas?

Answer 1

Answer: 296.17 K

Step-by-step explanation:

This question is a specific heat capacity question. Heat capacity is the proportionality constant which shows the relationship between the change in temperature and the heat,q.

It can represented mathematically by;

C= q/∆T------------------------------------***

STEP ONE:

Note that, ∆U= q---------------------------------------------------------------------------1).

DU= nCv∆T------------------------------(2).

PV= n×R×T (ideal gas equation)------------------------------------------------------(3).

Recall that ∆U= q; ∆U= nCv∆T. Substitute equation (3) into equation (2).

Hence, ∆U= (PV/RT) Cv∆T------------------------------------------------------------(4).

∆U= (PV/RT) Cv(T(f)- Ti). Where T(f)= final temperature and Ti= initial temperature.

Therefore, T(f )= Ti + RTiq/ PVCv -----------------------------------------------(5).

Slotting in the parameters given into equation (5).

T(f)= [(23+273.15)+ 8.314×2.66×10^4 × (23+273.15) ] ÷ 35×425×1.013×10^5× 20.8.

T(f)= 296.17 K.

Answer 2

75°C

Step-by-step explanation:

From the general gas equation: P1V1/T1 = P2V2/T2

P1 = 3.5atm = 3.5×101.325KN/m^2 = 354.64KN/m^2, V1 = 425L = 425/1000 = 0.425m^3, E1 = P1V1 = 354.64×0.425 = 150.722KJ, T1 = 23°C = 23 + 273K= 296K

E2 = P2V2 = 26.6 + E1 = 26.6 + 150.722 = 177.322KJ

T2 = (E2 × T1)/E1 = (177.322 × 296)/150.722 = 348K

T2 = 348 - 273 = 75°C

New temperature of the gas is 75°C