Respuesta :
The equation of state for an ideal gas is
[tex]pV=nRT[/tex]
where p is the gas pressure, V the volume, n the number of moles, R the gas constant and T the temperature.
The equation of state for the initial condition of the gas is
[tex]p_1 V_1 = n_1 R T_1[/tex] (1)
While the same equation for the final condition is
[tex]p_2 V_2 = n_2 R T_2[/tex] (2)
We know that in the final condition, half of the mass of the gas is escaped. This means that the final volume of the gas is half of the initial volume, and also that the final number of moles is half the initial number of moles, so we can write:
[tex]V_1 = 2 V_1[/tex]
[tex]n_1 = 2 n_2[/tex]
If we substitute these relationship inside (1), and we divide (1) by (2), we get
[tex] \frac{p_1}{p_2} = \frac{T_1}{T_2} [/tex]
And since the initial temperature of the gas is [tex]T_1 = 30 C=303 K[/tex], we can find the final temperature of the gas:
[tex]T_2 = T_1 \frac{p_2}{p_1}=(303 K) \frac{1.5 atm}{5.0 atm}=90.9 K [/tex]
[tex]pV=nRT[/tex]
where p is the gas pressure, V the volume, n the number of moles, R the gas constant and T the temperature.
The equation of state for the initial condition of the gas is
[tex]p_1 V_1 = n_1 R T_1[/tex] (1)
While the same equation for the final condition is
[tex]p_2 V_2 = n_2 R T_2[/tex] (2)
We know that in the final condition, half of the mass of the gas is escaped. This means that the final volume of the gas is half of the initial volume, and also that the final number of moles is half the initial number of moles, so we can write:
[tex]V_1 = 2 V_1[/tex]
[tex]n_1 = 2 n_2[/tex]
If we substitute these relationship inside (1), and we divide (1) by (2), we get
[tex] \frac{p_1}{p_2} = \frac{T_1}{T_2} [/tex]
And since the initial temperature of the gas is [tex]T_1 = 30 C=303 K[/tex], we can find the final temperature of the gas:
[tex]T_2 = T_1 \frac{p_2}{p_1}=(303 K) \frac{1.5 atm}{5.0 atm}=90.9 K [/tex]
Answer:
The final temperature of the tank is 181.8 K.
Explanation:
We use ideal gas equation to calculate the final temperature.
And volume remain constant for the given process.
Further Explanation:
The ideal gas equation is
PV= mRT
Here,
P is the pressure,
V is the volume,
T is the temperature
R is the gas constant
m is the mass of the gas.
The gas equation for the initial state can be written as
[tex]P_{1} V_{1} =m_{1} R T_{1}[/tex]
For final state gas equation becomes,
[tex][ P_{2} V_{2} =m_{2} R T_{2}[/tex]
During the process volume remain constant,
so [tex]V_{1} =V_{2} =[/tex]constant
From both the above equations, we get
[tex]\frac{P_{1} }{P_{2} } = \frac{m_{1}T_{1} }{m_{1}T_{2} }[/tex]
Given: [tex]m_{1} = 7kg, m_{2} =3.5 kg, P_{1} =5 atm, P_{2} = 1.5 atm[/tex] and [tex]T_{1} = 30^{0} C[/tex]
Substituting the given values, we get
[tex]\frac{5}{1.5}=\frac{(7)(303)}{(3.5)T_{2} }[/tex]
So final temperature,
[tex]T_{2}= 181.8 K[/tex]
Learn more:
https://brainly.in/question/2663978
Key word:
Ideal gas equation, Isochoric process.
