Question

For a mechanically reversible isobaric process taking place in a closed system involving 5 moles of an ideal gas, the temperature increases from an initial value of 300 K to a final value of 450 K. If the specific heat capacity at constant volume (Cv) is given as 12.5 J mol\(^{-1}\) K\(^{-1}\) and gas constant is 8.314 J mol\(^{-1}\) K\(^{-1}\), the amount of heat transferred to the system will be _ J. (Round off to the nearest integer)

Hint:

For isobaric processes, use \( Q = n C_p \Delta T \) to calculate the heat transferred, where \( C_p \) is related to \( C_v \) by \( C_p = C_v + R \).

Answer 1

Correct Answer: 15450

For an isobaric process, the heat transferred to the system \( Q \) can be calculated using the formula: \[ Q = n C_p \Delta T \] where: - \( n \) is the number of moles of the gas (given as 5 moles), - \( C_p \) is the specific heat capacity at constant pressure, - \( \Delta T \) is the change in temperature, i.e., \( T_{{final}} - T_{{initial}} \). We are given \( C_v = 12.5 \) J mol\(^{-1}\) K\(^{-1}\) and need to calculate \( C_p \), which is related to \( C_v \) by the equation: \[ C_p = C_v + R \] where \( R = 8.314 \) J mol\(^{-1}\) K\(^{-1}\) (the gas constant). Thus, \[ C_p = 12.5 + 8.314 = 20.814 \, {J mol}^{-1} {K}^{-1}. \] Now, we can calculate \( \Delta T \): \[ \Delta T = T_{{final}} - T_{{initial}} = 450 \, {K} - 300 \, {K} = 150 \, {K}. \] Substituting these values into the heat transfer equation: \[ Q = 5 \times 20.814 \times 150 = 15450 \, {J}. \] Thus, the amount of heat transferred to the system is \( \boxed{15450} \) J.