A sample of air weighing 1.18 g occupies 1.0 × 10^3 cm^3 when...

Given $P = 1.0 \times 10^5 \text{ Pa}$, $V = 1.0 \times 10^3 \text{ cm}^3 = 1.0 \times 10^{-3} \text{ m}^3$, and $T = 300 \text{ K}$.

Using the ideal gas equation $PV = nRT$, the value of $nR$ is:

$nR = \dfrac{PV}{T} = \dfrac{1.0 \times 10^5 \times 1.0 \times 10^{-3}}{300} = \dfrac{100}{300} = \dfrac{1}{3} \text{ J K}^{-1}$

The heat required at constant volume for \Delta T = 1^{}^{{}^{\circ}} \text{C} is $Q_v = 2.0 \text{ cal}$.

The heat required at constant pressure for the same temperature rise is $Q_p = Q_v + W$, where $W$ is the work done by the gas.

At constant pressure, $W = nR \Delta T$. For \Delta T = 1^{}^{{}^{\circ}} \text{C} = 1 \text{ K}:

$W = \dfrac{1}{3} \times 1 = \dfrac{1}{3} \text{ J}$

Converting the work into calories using the mechanical equivalent of heat $J = 4.2 \text{ J cal}^{-1}$ (noting $1 \text{ J} = 10^7 \text{ erg}$):

$W = \dfrac{1}{3 \times 4.2} \text{ cal} = \dfrac{1}{12.6} \text{ cal} \approx 0.079 \text{ cal}$

Therefore, the heat required at constant pressure is:

$Q_p = 2.0 + 0.079 = 2.079 \text{ cal} \approx 2.08 \text{ cal}$