A rigid tank contains 10 kg of an ideal gas at a pressure of 5 atm and a temperature of 30 oC. By opening a valve half of the mass of the gas is allowed to escape. In the final pressure in the tank is 1.8 atm, the final temperature in the tank, in oC, is

T₂ =  218.16 k = -54.84°C

Explanation:

Consider, the general gas equation:

PV = nRT

where,

P = Pressure of Gas

V = Volume of Gas

n = No. of Moles of Gas = mass in gram/Molecular Mass = m/M

R = Gas Constant

T = Temperature of Gas

Now,

PV = mRT/M

P/mT = RM/V

since, RM/V = Constant values in both final and initial states.

therefore,

P₁/m₁T₁ = P₂/m₂T₂

T₂ = P₂m₁T₁/m₂P₁

where,

T₂ = Final Temperature = ?

P₂ = Final Pressure = 1.8 atm

m₁ = Initial Mass = 10 kg

P₁ = Initial Pressure = 5 atm

m₂ = Final Mass = 10 kg/2 = 5 kg

T₁ = Initial Temperature = 30°C + 273 = 303 k

Therefore,

T₂ = (1.8 atm)(10 kg)(303 k)/(5 kg)(5 atm)

T₂ =  218.16 k = -54.84°C

c. newton's third law of motion

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