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 : 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:

where,

P = pressure of gas = ?

V = volume of gas =

T = temperature of gas =

R = gas constant =

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:

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.

or,

where,

= initial pressure of gas = 177.6 kPa

= final pressure of gas = 200 kPa

= initial temperature of gas =

= final temperature of gas = ?

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

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

a, c, d, e are all applying about success strategies.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.