First slide

Heat given

Question

Two moles of air, when heated through 10 K expands by an amount of 1.66 × 10^–3 m³ under a constant pressure of 10⁵ N/m². If C_v = 20.81 J/mole K, then C_p is,

Moderate

A

27.46J/mole K
B

19.11 J/mole K
C

32.11 J/mole k
D

19.11J/mole k

Solution

$\Delta W = P \times \Delta V = {10^5} \times 1.33 \times {10^{ - 3}}J$
$\Rightarrow \Delta W = 133J$
$\Delta U = n{C_V}\Delta T = 2 \times 20.81 \times 10J$
$\Rightarrow \Delta U = 416.2J$
Now, $\Delta Q = \Delta U + \Delta W$
= (416.2+133)J
= 529.2 J
Also, $\Delta Q = n{C_P}\Delta T$
$\therefore {C_P} = \frac{{\Delta Q}}{{n\Delta T}} = \frac{{549.2}}{{2 \times 10}}J/mol - K$
$= 27.46J/mol - K$

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