Question

Consider a sealed can that is filled with refrigerant-134a at 20 degrees C. Now a leak develops, and the pressure in the can drops to the local atmospheric pressure of 80 kPa. The temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer) Hint: The remaining refrigerant in the can contains both liquid and vapor at the final condition.

Answer 1

To calculate the final temperature of the refrigerant in the can, use the ideal gas law. The final temperature is 291 degrees C.

Step-by-step explanation:

To calculate the final temperature of the refrigerant in the can, we can use the ideal gas law. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

In this case, the volume and the number of moles of gas remain constant since the can is sealed. Therefore, we can rewrite the ideal gas law as P1/T1 = P2/T2.

Given that the initial pressure and temperature are 80 kPa and 20 degrees C, respectively, and the final pressure is 80 kPa, we can substitute those values into the equation to solve for the final temperature. Rearranging the equation, we get T2 = (P2/T1) * (P1).

Substituting the values, we have T2 = (80 kPa / (20 + 273)) * (80 kPa) = 291 K. Rounded to the nearest integer, the final temperature of the refrigerant in the can is 291 degrees C.

Answer 2

-29 degree

Step-by-step explanation:

Refrigerant in the can is expected to drop to -29^o C . The saturation temperature of R134a at 80kpa