Two moles of air, when heated through 10K expand by an amount of 1.66×10−3 m3 under a constant pressure of 105 N/m2 If Cv 20.81 J/mole Kthen Cp is

Cp−Cv=PdvndTCp−Cv=105×1.66×10−32×10=16.62=8.3 Cp=Cv+8.3=29.11 J/moleK

Two moles of air, when heated through 10K expand by an amount of 1.66×10−3 m3 under a constant pressure of 105 N/m2 If Cv 20.81 J/mole Kthen Cp is

Cp−Cv=PdvndTCp−Cv=105×1.66×10−32×10=16.62=8.3 Cp=Cv+8.3=29.11 J/moleK

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Two moles of air, when heated through 10K expand by an amount of $1.66 \times 10^{- 3} m^{3}$ under a constant pressure of $10^{5} N / m^{2}$ If C_v $20.81 J / m o l e K$ then C_p is

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44.34 J/mole K

38.21 J/mole K

39.11 J/mole K

29.11 J/mole K

answer is A.

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Detailed Solution

$C_{p} - C_{v} = \frac{P d v}{n d T}$

$C_{p} - C_{v} = \frac{10^{5} \times 1.66 \times 10^{- 3}}{2 \times 10} = \frac{16.6}{2} = 8.3$

$C_{p} = C_{v} + 8.3 = 29.11 J / m o l e K$

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Two moles of air, when heated through 10K expand by an amount of 1.66×10−3 m3 under a constant pressure of 105 N/m2 If Cv 20.81 J/mole Kthen Cp is

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Two moles of air, when heated through 10K expand by an amount of $1.66 \times 10^{- 3} m^{3}$ under a constant pressure of $10^{5} N / m^{2}$ If C_v $20.81 J / m o l e K$ then C_p is