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Heat and Thermodynamics question

2019 · 10 Apr · Shift 2 · Q66
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Heat and Thermodynamics question

2019 · 10 Apr · Shift 2 · Q66

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by Δ\DeltaΔ T. the heat required to produce the same change in temperature, at a constant pressure is :
  1. A
    75Q{7 \over 5}Q57​Q
  2. B
    32Q{3 \over 2}Q23​Q
  3. C
    23Q{2 \over 3}Q32​Q
  4. D
    53Q{5 \over 3}Q35​Q
View written solutionFree

Correct answer: A

  1. Use heat capacity relations

For an ideal gas,

  • At constant volume: QV=nCVΔTQ_V = n C_V \Delta TQV​=nCV​ΔT
  • At constant pressure: QP=nCPΔTQ_P = n C_P \Delta TQP​=nCP​ΔT

We are given that at constant volume, the supplied heat is QQQ, so Q=nCVΔTQ = n C_V \Delta TQ=nCV​ΔT

We need the heat required at constant pressure for the same temperature rise ΔT\Delta TΔT: QP=nCPΔTQ_P = n C_P \Delta TQP​=nCP​ΔT

  1. Heat capacities for a diatomic rigid molecule gas

For a diatomic gas of rigid molecules, the degrees of freedom are:

  • 3 translational
  • 2 rotational

So total degrees of freedom: f=5f=5f=5

Hence, CV=f2R=52RC_V = \frac{f}{2}R = \frac{5}{2}RCV​=2f​R=25​R

and CP=CV+R=52R+R=72RC_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}RCP​=CV​+R=25​R+R=27​R

  1. Find the ratio

QPQV=nCPΔTnCVΔT=CPCV=72R52R=75\frac{Q_P}{Q_V} = \frac{n C_P \Delta T}{n C_V \Delta T} = \frac{C_P}{C_V} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}QV​QP​​=nCV​ΔTnCP​ΔT​=CV​CP​​=25​R27​R​=57​

Since QV=QQ_V = QQV​=Q, QP=75QQ_P = \frac{7}{5}QQP​=57​Q

  1. Match with options

The correct option is: A: 75Q\boxed{\text{A: } \frac{7}{5}Q}A: 57​Q​

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