The tank contains 2 kg of air at an absolute pressure of 400 kpa and a temperature of 20°c. if 0.6 kg of air is added to the tank and the temperature rises to 32°c, determine the pressure in the tank.

The pressure in tank will become 541.3 KPa.

Explanation:

From general gas equation we know that:

PV = nRT

PV = (m/M)RT

P/Tm = R/VM

Now, for initial state:

P1/T1 m1 = R/VM eqn (1)

For Final State, as the gas constant (R), volume (V) and molecular mass of air (M) remains constant;

P2/T2 m2 = R/VM eqn (2)

Comparing eqn (1) and eqn (2):

P1/T1 m1 = P2/T2 m2

Now, we have:

P1 = 400 KPa

P2 = ?

m1 = 2 kg

m2 = 0.6 kg + 2 kg = 2.6 kg

T1 = 20°C = 293 K

T2 = 32°C = 305 K

Therefore,

(400 KPa)/(293 K)(2 kg) = P2/(305 K)(2.6 kg)

P2 = 541.3 KPa

can you make it more obvious. i can't see it properly