Question

The molar specific heat of oxygen at constant pressure $Cp=7.03$ cal/mol and $R=8.31$ J/mol. Heat taken by 5 moles from $10^\circC$ to $20^\circC$ at constant volume is approximately:

• A. 25 cal
• B. 50 cal
• C. 253 cal
• D. 500 cal

Hint:

$Cp - Cᵥ = R$ (in Joules). $1$ cal = $4.18$ J.

Answer 1

The Correct Option is C

To find the heat taken by 5 moles of oxygen when the temperature is increased from \(10^\circ C\) to \(20^\circ C\) at constant volume, we need to use the concept of specific heat at constant volume (\(C_v\)). The molar specific heat at constant volume can be calculated using the equation \(C_v = C_p - R\), where \(C_p\) is the molar specific heat at constant pressure and \(R\) is the universal gas constant.

1. We know:
   • Molar specific heat at constant pressure, \(C_p = 7.03 \text{ cal/mol/}^\circ C\)
   • Number of moles, \(n = 5\)
   • Initial temperature, \(T_1 = 10^\circ C\)
   • Final temperature, \(T_2 = 20^\circ C\)
   • Universal gas constant, \(R = 8.31 \text{ J/mol/}^\circ C\)
   • Conversion: \(1 \text{ cal} = 4.184 \text{ J}\)
2. Convert \(R\) from Joules to calories:

\(R = \frac{8.31}{4.184} \approx 1.99 \text{ cal/mol/}^\circ C\)

1. Calculate molar specific heat at constant volume:

\(C_v = C_p - R = 7.03 - 1.99 = 5.04 \text{ cal/mol/}^\circ C\)

1. Use the formula for heat transfer at constant volume: \(Q = nC_v\Delta T\), where \(\Delta T = (T_2 - T_1)\) is the change in temperature.

\(\Delta T = 20 - 10 = 10^\circ C\)

1. Substitute the values to find \(Q\):

\(Q = 5 \times 5.04 \times 10 = 252 \text{ cal}\)

This result is approximately 253 cal. Therefore, the correct answer is approximately 253 cal.

1. Conclusion: The heat taken by 5 moles of oxygen from \(10^\circ C\) to \(20^\circ C\) at constant volume is approximately 253 cal.