Question

The heat to be supplied to 20 g of oxygen to increase its temperature from 27 $^\circ$C to 59 $^\circ$C at constant pressure is (Universal gas constant = 8.3 J mol$^{-1}$ K$^{-1}$)

• A. 1162 J
• B. 423 J
• C. 934 J
• D. 581 J

Hint:

When dealing with gases, always remember the specific heat capacities for different types of gases (monatomic, diatomic, polyatomic). For a diatomic gas at constant pressure, $C_p = \frac{7}{2}R$. Ensure consistent units (SI units) throughout the calculation.

Answer 1

The Correct Option is D

Step 1: Convert given temperatures to Kelvin and calculate the change in temperature.
Initial temperature, $T_1 = 27^\circ \text{C} = 27 + 273.15 \text{ K} = 300.15 \text{ K}$ Final temperature, $T_2 = 59^\circ \text{C} = 59 + 273.15 \text{ K} = 332.15 \text{ K}$ Change in temperature, $\Delta T = T_2 - T_1 = 332.15 \text{ K} - 300.15 \text{ K} = 32 \text{ K}$.
Alternatively, $\Delta T = 59^\circ \text{C} - 27^\circ \text{C} = 32^\circ \text{C}$, which is equivalent to $32 \text{ K}$ for temperature difference. Step 2: Calculate the number of moles (n) of oxygen.
Oxygen is a diatomic gas ($O_2$). Its molar mass ($M$) is $2 \times 16 \text{ g/mol} = 32 \text{ g/mol}$.
Given mass of oxygen, $m = 20 \text{ g}$.
Number of moles, $n = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \text{ g}}{32 \text{ g/mol}} = \frac{5}{8} \text{ mol}$. Step 3: Determine the molar heat capacity at constant pressure ($C_p$) for oxygen.
For a diatomic gas like oxygen, the molar specific heat at constant volume ($C_v$) is $\frac{5}{2}R$ (considering 3 translational and 2 rotational degrees of freedom).
The molar specific heat at constant pressure ($C_p$) is related to $C_v$ by Mayer's relation: $C_p = C_v + R$.
\[ C_p = \frac{5}{2}R + R = \frac{7}{2}R \] Given Universal gas constant, $R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$.
\[ C_p = \frac{7}{2} \times 8.3 \text{ J mol}^{-1} \text{ K}^{-1} = 3.5 \times 8.3 \text{ J mol}^{-1} \text{ K}^{-1} = 29.05 \text{ J mol}^{-1} \text{ K}^{-1} \] Step 4: Calculate the heat supplied ($Q_p$) at constant pressure.
The heat supplied at constant pressure is given by:
\[ Q_p = n C_p \Delta T \] Substitute the calculated values:
\[ Q_p = \left(\frac{5}{8} \text{ mol}\right) \times \left(29.05 \text{ J mol}^{-1} \text{ K}^{-1}\right) \times (32 \text{ K}) \] \[ Q_p = \frac{5}{8} \times 29.05 \times 32 \] \[ Q_p = 5 \times 29.05 \times 4 \] \[ Q_p = 20 \times 29.05 \] \[ Q_p = 581 \text{ J} \] The final answer is $\boxed{581 \text{ J}}$.