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## Q. 2.23

In a vessel 10 kg of air is heated in a reversible, non-flow, constant volume process so that pressure of air is increased two times that of the initial value. The initial temperature is $20^0$C. Calculate (a) the final temperature, (b) change in internal energy, (c) change in enthalpy, and (d) heat transfer. Take R = 0.287 kJ/kg. K and $C_ν$ = 0.718 kJ/kg. K for air.

## Verified Solution

For constant volume process:

(a) $T_2=\left(p_2 / p_1\right) T_1=2 \times 293=586 K \text { or } 313^0 C$

(b) $d U=m C_ν\left(T_2-T_1\right)=10 \times 0.718(586-293)=2103.74 \ kJ$

(c) $d H=m C_p\left(T_2-T_1\right)=10 \times(0.718+0.287)(586-293)=2944.65 \ kJ$

From First law of thermodynamics,

$Q_{1-2}=d U+W_{1-2}$

For constant volume process, $W_{1-2}=0$

$\therefore \quad \quad Q_{1-2}=d U=2103.74 \ kJ$