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70 cal of heat are required to raise the temperature of 2 moles of an ideal diatomic gas at constant pressure from 30ºC to 35°C. The amount of heat required (in calorie) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume is
  • a)
    50
  • b)
    90
  • c)
    70
  • d)
    30
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
70 cal of heat are required to raise the temperature of 2 moles of an ...
We know that heat required to raise the temperature at constant pressure is Cp.
So,  
Similarly, at constant volume CV

or 
The correct answer is: 50
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Most Upvoted Answer
70 cal of heat are required to raise the temperature of 2 moles of an ...
Assuming the ideal diatomic gas is behaving according to the ideal gas law, we can use the equation:

q = n * C * ΔT

Where:
q = heat transfer (in calories)
n = number of moles of gas
C = molar heat capacity at constant pressure (in calories/mol °C)
ΔT = change in temperature (in °C)

In this case, we are given that 70 cal of heat are required to raise the temperature of 2 moles of the gas at constant pressure. Therefore, we have:

q = 70 cal
n = 2 moles

We need to find ΔT, so we rearrange the equation:

ΔT = q / (n * C)

Since the heat capacity at constant pressure for an ideal diatomic gas is 7/2 R (where R is the ideal gas constant), we have:

C = 7/2 R

Now we can substitute the given values and solve for ΔT:

ΔT = 70 cal / (2 moles * (7/2 R))

ΔT = 70 cal / (2 moles * (7/2 * 0.0821 L atm/mol K))

ΔT = 70 cal / (2 moles * 0.287 L atm/mol K)

ΔT ≈ 121.95 K

Therefore, the temperature of the gas increases by approximately 121.95 °C.
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70 cal of heat are required to raise the temperature of 2 moles of an ideal diatomic gas at constant pressure from 30ºC to 35°C. The amount of heat required (in calorie) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume isa)50b)90c)70d)30Correct answer is option 'A'. Can you explain this answer?
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