Answer:
The value is [tex]m_2 = 2.46 \ kg[/tex]
Explanation:
From the question we are told that
The volume of the tank is [tex]V = 0.57 \ m^3[/tex]
The mass of the Argon it contains is [tex]m = 4 \ kg[/tex]
The initial pressure on of the gas is [tex]P_1 = 450 kPa = 450 *10^{3} \ Pa[/tex]
The initial temperature is [tex]T_1 = 30^oC = 30 + 273 = 303 \ K[/tex]
The new pressure inside the tank is [tex]P_2 = 200 \ kPa = 200 *10^{3} \ Pa[/tex]
Gnerally given that the Argon remaining inside the tank has undergone a reversible, adiabatic process, then the final temperature of the Argon gas is mathematically represented as
[tex]T_2 = T_1 * [\frac{P_2}{P_1} ]^{ \frac{(k - 1 )}{k} }[/tex]
Here k is the specific heat ratio of Argon with value [tex]k = 1.667[/tex]
So
[tex]T_2 = 303 * [\frac{200}{450} ]^{ \frac{(1.667- 1 )}{1.667} }[/tex]
=> [tex]T_2 = 219 \ K[/tex]
Generally from the ideal gas equation
[tex]PV = mRT_2[/tex]
So
[tex]\frac{P_1V}{P_2V} =\frac{ m_1 * R * T_1 }{ m_2 * R * T_2}[/tex]
Here R is the gas constant with value [tex]R = 8.314 \J \ K^{-1}\ mol^{-1}[/tex]
=> [tex]m_2 = \frac{P_2 * T_1 }{ P_1 * T_2 } * m_1[/tex]
=> [tex]m_2 = \frac{200 * 303 }{ 450 * 219 } * 4[/tex]
=> [tex]m_2 = 2.46 \ kg[/tex]
