Answer:
(A) 10132.5Pa
(B)531kJ of energy
Explanation:
This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.
Given
m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³
P1 = 101325Pa. M = 102.03g/mol
P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa
(B) Energy is transfered by the r134a in the form of thw work done in in expansion
W = nRTIn(V2/V1)
n = m / M = 10000/102.03 = 98.01mols
W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)
= 531kJ.