Question

10 kg of R-134a fill a 1.115-m^3 rigid container at an initial temperature of -30∘C. The container is then heated until the pressure is 200 kPa. Determine the final temperature an the initial pressure. Answers: 14.2∘C, 84.43 kPa

Answer 1

Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

Explanation :

First we have to calculate the initial pressure of gas.

As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of gas = ?

V = volume of gas = $1.115m^3$

T = temperature of gas = $-30^oC=273+(-30)=243K$

R = gas constant = $8.314m^3Pa/mole.K$

w = mass of gas = 10 kg = 10000 g

M = molar mass of R-134a gas = 102.03 g/mole

Now put all the given values in the ideal gas equation, we get:

$P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)$

$P=177587.9687Pa=177.6kPa$

Thus, the initial pressure of gas is, 177.6 kPa

Now we have to calculate the final temperature of gas.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 177.6 kPa

$P_2$ = final pressure of gas = 200 kPa

$T_1$ = initial temperature of gas = $-30^oC=273+(-30)=243K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{177.6kPa}{243K}=\frac{200kPa}{T_2}$

$T_2=273.6K$

Thus, the final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa