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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Thermodynamics

BITSAT 2024 · Physics · Thermodynamics

In the given cycle A B C D A , the heat required for an ideal...

In the given cycle ABCDAA B C D A, the heat required for an ideal monoatomic gas will be In the given cycle A B C D A , the heat required for an ideal monoatomic gas will be

Options

  1. Ap0V0p_0 V_0
  2. B132p0V0\frac{13}{2} p_0 V_0
  3. C112p0V0\frac{11}{2} p_0 V_0
  4. D4p0V04 p_0 V_0

Correct answer

B. 132p0V0\frac{13}{2} p_0 V_0

Step-by-step solution

In the given cycle, heat is supplied in process DD to AatconstantvolumeandinprocessAtoBat constant volume and in process A to B at constant pressure. ∴\therefore Heat supplied, Q=nCV(ΔT)DA+nCp(ΔT)ABQ=n C_V(\Delta T)_{D A}+n C_p(\Delta T)_{A B} For an ideal monoatomic gas, CV=32R and Cp=52R⇒Q=32nRΔT+52nRΔT[∵nRΔT=pΔV]=32(p0V0)+5(p0V0)=132p0V0\begin{aligned} C_V & =\frac{3}{2} R \text{ and } C_p=\frac{5}{2} R \\ \Rightarrow \quad Q & =\frac{3}{2} n R \Delta T+\frac{5}{2} n R \Delta T \quad[\because n R \Delta T=p \Delta V] \\ & =\frac{3}{2}\left(p_0 V_0\right)+5\left(p_0 V_0\right) \\ & =\frac{13}{2} p_0 V_0 \end{aligned}
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