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kumpel [21]
3 years ago
14

A rigid tank contains 7 kg of an ideal gas at 5 atm and 30c. a valve is opened, and half of mass of the gas can escape. the fin

al pressure in the tank is 1.5 atm.calculate the final temperature in the tank.
Physics
2 answers:
Kruka [31]3 years ago
8 0

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  

P_{1} V_{1} =m_{1} R T_{1}  

For final state gas equation becomes,

[ P_{2} V_{2} =m_{2} R T_{2}

During the process volume remain constant,

so V_{1} =V_{2} =constant

From both the above equations, we get

\frac{P_{1} }{P_{2} } = \frac{m_{1}T_{1} }{m_{1}T_{2} }

Given: m_{1} = 7kg, m_{2}  =3.5 kg, P_{1} =5 atm, P_{2} = 1.5 atm  and T_{1} = 30^{0} C

Substituting the given values, we get

\frac{5}{1.5}=\frac{(7)(303)}{(3.5)T_{2} }

So final temperature,

T_{2}=  181.8 K

Learn more:

https://brainly.in/question/2663978

Key word:

Ideal gas equation, Isochoric process.  

Readme [11.4K]3 years ago
6 0
The equation of state for an ideal gas is
pV=nRT
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
p_1 V_1 = n_1 R T_1 (1)
While the same equation for the final condition is
p_2 V_2 = n_2 R T_2 (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:
V_1 = 2 V_1
n_1 = 2 n_2
If we substitute these relationship inside (1), and we divide (1) by (2), we get
\frac{p_1}{p_2} = \frac{T_1}{T_2}

And since the initial temperature of the gas is T_1 = 30 C=303 K, we can find the final temperature of the gas:
T_2 = T_1  \frac{p_2}{p_1}=(303 K) \frac{1.5 atm}{5.0 atm}=90.9 K
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