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

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., \( T_{\text{final}} - T_{\text{initial}} \).

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, we have:
\[ C_p = 12.5 + 8.314 = 20.814 \, \text{J mol}^{-1} \text{K}^{-1} \].

Now, we can calculate \( \Delta T \):
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 450 \, \text{K} - 300 \, \text{K} = 150 \, \text{K} \].

Substituting these values into the heat transfer equation:
\[ Q = 5 \times 20.814 \times 150 = 15450 \, \text{J} \].

Thus, the amount of heat transferred to the system is \( \boxed{15450} \) J.