ANS:C - 18.11 kJ

To find the heat given off by 1 mole of air when cooled at atmospheric pressure from 500°C to -100°C, we can use the equation for heat capacity provided: 𝐶𝑝=26.693+7.365×10−3𝑇Cp​=26.693+7.365×10−3T Where:

• 𝐶𝑝Cp​ is the heat capacity in J/mole.K
• 𝑇T is the temperature in Kelvin

We need to integrate this equation with respect to temperature (𝑇T) from 500°C to -100°C (which need to be converted to Kelvin). The heat given off is equal to the change in enthalpy (Δ𝐻ΔH) of the air, which can be calculated using the formula: Δ𝐻=∫𝑇1𝑇2𝐶𝑝 𝑑𝑇ΔH=∫T1​T2​​Cp​dT Where:

• 𝑇1T1​ is the initial temperature (in Kelvin)
• 𝑇2T2​ is the final temperature (in Kelvin)
• 𝐶𝑝Cp​ is the heat capacity function

First, we convert the temperatures from Celsius to Kelvin: 𝑇1=500°𝐶=500+273.15 K=773.15 KT1​=500°C=500+273.15K=773.15K 𝑇2=−100°𝐶=−100+273.15 K=173.15 KT2​=−100°C=−100+273.15K=173.15K Now, we integrate the given heat capacity equation with respect to temperature from 𝑇1T1​ to 𝑇2T2​: Δ𝐻=∫773.15173.15(26.693+7.365×10−3𝑇) 𝑑𝑇ΔH=∫773.15173.15​(26.693+7.365×10−3T)dT Δ𝐻=[26.693𝑇+7.3652×10−3𝑇2]773.15173.15ΔH=[26.693T+27.365​×10−3T2]773.15173.15​ Δ𝐻=[26.693×173.15+7.3652×10−3×(173.15)2]−[26.693×773.15+7.3652×10−3×(773.15)2]ΔH=[26.693×173.15+27.365​×10−3×(173.15)2]−[26.693×773.15+27.365​×10−3×(773.15)2] Δ𝐻≈(4616.677+90.953)−(20642.827+8934.529)ΔH≈(4616.677+90.953)−(20642.827+8934.529) Δ𝐻≈4707.63 J/molΔH≈4707.63J/mol Finally, we convert this value to kJ/mol: Δ𝐻≈4707.631000 kJ/molΔH≈10004707.63​kJ/mol Δ𝐻≈4.708 kJ/molΔH≈4.708kJ/mol So, the heat given off by 1 mole of air when cooled at atmospheric pressure from 500°C to -100°C is approximately 4.708 kJ4.708kJ. Among the provided options, none exactly matches this value. However, the closest option is 4.73 kJ. Therefore, 4.73 kJ is the closest option to the calculated value.