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70 calories of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30°C to 35°C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is
We know that, Q = nCpdT… (i),
Where
Q = heat required to raise the temperature
n = number of moles
Cp = specific heat capacity at constant pressure
dT = rise in temperature.
In this first case,
Amount of heat required(Q) = 70 cal
Number of moles(n) = 2 mol
Rise in temperature(dT) = (35-30)0C = 50C
Hence, from (i), we get Cp = 7 calmol-1 0C-1
Now, we know, Cp - Cv = R,
Where
Cp = specific heat capacity at constant pressure
Cv = specific heat capacity at constant volume
R = universal gas constant = 1.98 calmol-1 0C-1
Therefore, we get Cv = Cp - R = (7-1.98) cal mol-1 0C-1
= 5.02 cal mol-1 0C-1
Now, in the new case, change in temperature dT = 50C as before. Number of moles(n) = 2, and Cv = 5.02 calmol-1 0C-1
Hence, amount of heat required to raise the temperature of the same gas through the same range
dQ = nCvdT = (2 x 5.02 x 5) cal = 50.2 cal which is approximately equal to 50 cal.
Hence, the correct option is (b).