32 gram of oxygen gas at temperature 27 ° C is compressed adiabatically to 1 3 of its initial volume. Calculate the change in internal energy ( γ = 1 . 5 for oxygen).

For the process to take place adiabatically, we know that internal energy is given as, d U = n C v d T = n R ( γ − 1 ) d T . Now, P V γ = constant . So, we can say that, V γ × n R T V = constant ⇒ T V ( γ − 1 ) = constant . So, we can say, T 1 V 1 ( γ − 1 ) = T 2 V 2 ( γ − 1 ) 300 V 1 ( γ − 1 ) = T 2 V 3 ( γ − 1 ) 300 T 2 = 1 3 1 . 5 - 1 T 2 = 300 1 / 3 = 300 3 K N = number of moles = 32 32 = 1 . Now, the internal energy can be written as, d U = 1 × R ( 1 . 5 − 1 ) × 300 1 . 732 - 1 ⇒ 2 R × 300 3 − 1 = 2 × 25 3 × 300 1 . 732 − 1 = 50 × 100 × 0 . 732 J = 3660 J .

The coefficient of performance of a carnot refrigerator is 5 3 . If the refrigerator operates at an intake temperature of - 23 ° C , what will be the exhaust temperature?

A refrigerator works in a reverse Carnot cycle. It has three main components: the source, the sink and the working substance. Its performance is analysed by the coefficient of performance which is the ratio of heat released to the sink to the work done to release the heat. Coefficient of performance for a carnot engine is given by, C O P = T low T high − T low , ⇒ 5 3 = 273 - 23 T high - 273 - 23 ⟹ T high − 250 = 150 ⟹ T high = 400 K = 127 ° C .