One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer?

Question

One kg of air, initially at a temperature of 127&deg;C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).Correct answer is '114.8'. Can you explain this answer?

Answer

°C, is heated at constant pressure until its temperature reaches 227°C. We are asked to find the change in heat transfer during this process.

To solve this problem, we can use the equation:

Q = mcΔT

Where:
Q = heat transfer
m = mass of the substance (in this case, air) = 1 kg
c = specific heat capacity of air at constant pressure = 1.005 kJ/kg°C (given)
ΔT = change in temperature = final temperature - initial temperature

First, we need to convert the temperatures from Celsius to Kelvin:
Initial temperature (Ti) = 127°C + 273.15 = 400.15 K
Final temperature (Tf) = 227°C + 273.15 = 500.15 K

Now, we can calculate the change in temperature:
ΔT = Tf - Ti
ΔT = 500.15 K - 400.15 K
ΔT = 100 K

Substituting the values into the equation, we get:
Q = (1 kg)(1.005 kJ/kg°C)(100 K)
Q = 100.5 kJ

Therefore, the change in heat transfer during this process is 100.5 kJ.

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