Can someone help me with these 2 problems thanks 120 kPa A r
Can someone help me with these 2 problems thanks
 Solution
The container is rigid.So, the process is isochoric process.
(b).No.of moles n = 11.3 mol
Initial pressure P = 61 kPa = 61 x10 3 Pa
Final pressure P \' = 120 kPa = 120 x10 3 Pa
Initial temprature T = 381.58 o C = 381.58 + 273 = 654.58 K
In Isochoric process, P \' / P = T \' / T
From this final pressure T \' = ( P \' / P ) T
= (120 kPa / 61 kPa) 654.58 K
= 1287.69 K
= (1287.69 - 273 ) o C
= 1014.69 o C

