Sol:

For isobaric process,

$C_{p} = \frac{\Delta Q}{n\Delta T}$ $C_{p} = \frac{70}{2 * (35 – 30)}$ $C_{p} = \frac{70}{2 * 5}$ $C_{p} = \frac{70}{10} = 7 calories mol^{-1}K^{-1}$

For an idea; gas,

$C_{p} – C_{v} = R = 8.314J-mol^{-1}K^{-1}\simeq 2 calories mol^{-1}K^{-1}$ $\Rightarrow C_{v} = C_{p} – R$ $\Rightarrow C_{v} = (7 – 2)$ $\Rightarrow C_{v} = 5 calories mol^{-1}K^{-1}$ $\Rightarrow C_{v} = \frac{\Delta Q}{n\Delta T}$ $\Rightarrow 5 = \frac{\Delta Q}{2 * (35 – 30)}$ $\Rightarrow \Delta Q = 5 * 2 * (35 – 30)$ $\Rightarrow \Delta Q = 5 * 2 * 5$ $\Rightarrow \Delta Q = 50$

Therefore, 50 calories of heat are required to raise the temperature of two moles of

gas from 30° – 35° C at a constant volume.

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