JEE Challenger
More from Thermodynamics
PhyHardAdv 2016 • Paper 2

Heat Supplied in Two Step Thermodynamic Process

A gas is contained in a cylinder equipped with a frictionless movable piston. Its initial state defined by Pi=105 PaP_i = 10^5 \text{ Pa} and volume Vi=10−3 m3V_i = 10^{-3} \text{ m}^3 transforms to a final state with Pf=(1/32)×105 PaP_f = (1/32) \times 10^5 \text{ Pa} and Vf=8×10−3 m3V_f = 8 \times 10^{-3} \text{ m}^3 via a quasi-static adiabatic process following P3V5=constantP^3 V^5 = \text{constant}. Now consider another path bringing the gas from the identical initial state to the same final state via two distinct steps: an isobaric expansion at PiP_i followed by an isochoric process at VfV_f. Calculate the approximate amount of heat supplied to the system in this alternate two-step process.

Options

A

112 J112 \text{ J}

B

294 J294 \text{ J}

C

588 J588 \text{ J}

Correct
D

813 J813 \text{ J}

Topics & Concepts

ThermodynamicsFirst Law of ThermodynamicsHeat Transfer

Step-by-Step Solution

To find the amount of heat supplied to the gas in the alternate two-step process, we use the First Law of Thermodynamics: Q=ΔU+WQ = \Delta U + W

Step 1: Determine the Degrees of Freedom and Internal Energy Change (ΔU\Delta U)

The quasi-static adiabatic process follows the relation: P3V5=constantP^3 V^5 = \text{constant}

Taking the cube root on both sides: PV5/3=constantP V^{5/3} = \text{constant}

Comparing this with the adiabatic equation PVγ=constantP V^{\gamma} = \text{constant}, we get: γ=53\gamma = \frac{5}{3}

This corresponds to a monoatomic ideal gas, for which the molar heat capacity at constant volume is: Cv=32RC_v = \frac{3}{2}R

Since internal energy UU is a state function, the change in internal energy ΔU\Delta U depends only on the initial and final states: ΔU=nCvΔT=32(PfVf−PiVi)\Delta U = n C_v \Delta T = \frac{3}{2} (P_f V_f - P_i V_i)

Given values:

  • Pi=105 PaP_i = 10^5 \text{ Pa}
  • Vi=10−3 m3V_i = 10^{-3} \text{ m}^3
  • Pf=132×105 PaP_f = \frac{1}{32} \times 10^5 \text{ Pa}
  • Vf=8×10−3 m3V_f = 8 \times 10^{-3} \text{ m}^3

Calculating PiViP_i V_i and PfVfP_f V_f: PiVi=105×10−3=100 JP_i V_i = 10^5 \times 10^{-3} = 100 \text{ J} PfVf=(132×105)×(8×10−3)=832×100=25 JP_f V_f = \left(\frac{1}{32} \times 10^5\right) \times (8 \times 10^{-3}) = \frac{8}{32} \times 100 = 25 \text{ J}

Now, calculate ΔU\Delta U: ΔU=32(25−100)=32(−75)=−112.5 J\Delta U = \frac{3}{2} (25 - 100) = \frac{3}{2} (-75) = -112.5 \text{ J}

Step 2: Calculate the Work Done (WW) in the Two-Step Process

The alternate process consists of two steps:

  1. Isobaric expansion at pressure PiP_i from ViV_i to VfV_f: W1=Pi(Vf−Vi)=105×(8×10−3−10−3)=105×7×10−3=700 JW_1 = P_i (V_f - V_i) = 10^5 \times (8 \times 10^{-3} - 10^{-3}) = 10^5 \times 7 \times 10^{-3} = 700 \text{ J}

  2. Isochoric cooling at constant volume VfV_f from PiP_i to PfP_f: W2=0 JW_2 = 0 \text{ J}

Total work done in this path: W=W1+W2=700 JW = W_1 + W_2 = 700 \text{ J}

Step 3: Calculate the Heat Supplied (QQ)

Using the First Law of Thermodynamics: Q=ΔU+W=−112.5 J+700 J=587.5 J≈588 JQ = \Delta U + W = -112.5 \text{ J} + 700 \text{ J} = 587.5 \text{ J} \approx 588 \text{ J}

Thus, the correct option is (C).

PreviousNext