Respuesta :
Answer:
The correct answer is option c.
Explanation:
Using An deal gas equation:
[tex]PV=nRT[/tex]
V = Volume of the gas at given pressure P and Temperature T.
n= number of moles
[tex]n=\frac{m}{M}[/tex]
m = Mass of the gas
M= Molar mass of the gas
[tex]\frac{PVM}{mT}=R= constant[/tex]
Volume of the gas at 25 kPa and 120°C,[tex]V_1 = 4 m^3[/tex]
[tex]P_1=25 kPa+ 101.3 kpa=126.3 kpa[/tex]
(Absolute pressure is equal to sum of gauge pressure ant atmospheric pressure.)
[tex]T_1=120^oC=393.15 K[/tex]
Moles of gas ,[tex]n_1=\frac{m_1}{M}[/tex]
[tex]\frac{P_1V_1M}{m_1T_1}=R[/tex]...(1)
Volume of the gas at 101.3 kPa and 20°C,[tex]V_2= ? m^3[/tex]
[tex]P_2=101.3 kpa,T_2=20^oC=293.15 K[/tex]
Moles of gas ,[tex]n_2=\frac{m_2}{M}[/tex]
Volume asked for the same mass of gas:
[tex]m_1=m_2[/tex]
[tex]\frac{P_1V_1M}{m_2T_1}=R[/tex]..(2)
[tex]\frac{P_1V_1M}{m_2T_1}=\frac{P_2V_2M}{m_2T_2}=R[/tex]
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
Substituting all the given values we get value of [tex][V_2[/tex]:
[tex]V_2=3.72 m^3[/tex]
