Inverse: Q in an isobaric process

Problem

One mole of diatomic gas at $T_1 = 300\ \text{K}$ absorbs $Q = 4000\ \text{J}$ at constant pressure. Calculate the final temperature, using $c_p = \tfrac{7}{2}R$.

Solution

Heat at constant pressure.

$Q = n\,c_p\,\Delta T$

Inversion. With $n = 1$ and $c_p = \tfrac{7}{2}R = \tfrac{7}{2}\cdot 8{,}31 = 29{,}085\ \text{J/(mol·K)}$:

$\Delta T = \frac{Q}{n\,c_p} = \frac{4000}{29{,}085} = 137{,}5\ \text{K}$

Final temperature.

$T_2 = T_1 + \Delta T = 300 + 137{,}5 \approx 437{,}5\ \text{K}$

Conclusion.

$\ev{T_2\approx 437\ \text{K}}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes · Specific heat and heat capacity Objects: Ideal gas