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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 = nC_{p}dT… (i),

Where

Q = heat required to raise the temperature

n = number of moles

C_{p} = 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)^{0}C = 5^{0}C

Hence, from (i), we get C_{p} = 7 calmol^{-1 0}C^{-1}

Now, we know, C_{p} - C_{v} = R,

Where

C_{p} = specific heat capacity at constant pressure

C_{v} = specific heat capacity at constant volume

R = universal gas constant = 1.98 calmol^{-1 0}C^{-1}

Therefore, we get C_{v} = C_{p} - R = (7-1.98) cal mol^{-1 0}C^{-1}

= 5.02 cal mol^{-1 0}C^{-1}

Now, in the new case, change in temperature dT = 5^{0}C as before. Number of moles(n) = 2, and C_{v} = 5.02 calmol^{-1 0}C^{-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).

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