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A monoatomic gas at a pressure PP, having a volume VV, expands isothermally to a volume 2V2V and then adiabatically to a volume 16V16V. The final pressure of the gas is: (Take: γ=53\gamma = \frac{5}{3})

A

64P64P

B

32P32P

C

P64\frac{P}{64}

D

16P16P

Step-by-Step Solution

The process consists of two steps:

Step 1: Isothermal Expansion Initial state: P1=P,V1=VP_1 = P, V_1 = V Final state of step 1: V2=2VV_2 = 2V For an isothermal process, PV=constantPV = \text{constant} (Boyle's Law). P1V1=P2V2P_1 V_1 = P_2 V_2 PV=P2(2V)P \cdot V = P_2 \cdot (2V) P2=P2P_2 = \frac{P}{2}

Step 2: Adiabatic Expansion Initial state of step 2: P2=P2,V2=2VP_2 = \frac{P}{2}, V_2 = 2V Final state: V3=16VV_3 = 16V For an adiabatic process, PVγ=constantPV^{\gamma} = \text{constant}. P2V2γ=P3V3γP_2 V_2^{\gamma} = P_3 V_3^{\gamma} Substituting the values: (P2)(2V)γ=P3(16V)γ(\frac{P}{2}) (2V)^{\gamma} = P_3 (16V)^{\gamma} P3=P2(2V16V)γP_3 = \frac{P}{2} \left( \frac{2V}{16V} \right)^{\gamma} P3=P2(18)53P_3 = \frac{P}{2} \left( \frac{1}{8} \right)^{\frac{5}{3}} P3=P2((12)3)53P_3 = \frac{P}{2} \left( (\frac{1}{2})^3 \right)^{\frac{5}{3}} P3=P2(12)5P_3 = \frac{P}{2} \left( \frac{1}{2} \right)^{5} P3=P2132=P64P_3 = \frac{P}{2} \cdot \frac{1}{32} = \frac{P}{64}

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