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Question: P/V^2= constant given the 1mole gas is expanded from 1 to 2 L initial temp 100K find work done in c...

P/V^2= constant given the 1mole gas is expanded from 1 to 2 L initial temp 100K find work done in calories given R=0.0821 and 1Latm=24 cal

Answer

-459.76 cal

Explanation

Solution

The problem asks us to calculate the work done during the expansion of 1 mole of gas under a specific condition (P/V2=constantP/V^2 = \text{constant}).

1. Determine the initial pressure (P1P_1) Given: Number of moles (nn) = 1 mol Initial volume (V1V_1) = 1 L Initial temperature (T1T_1) = 100 K Gas constant (RR) = 0.0821 L atm K−1^{-1} mol−1^{-1}

Using the ideal gas equation, P1V1=nRT1P_1V_1 = nRT_1: P1=nRT1V1=1 mol×0.0821 L atm K−1 mol−1×100 K1 LP_1 = \frac{nRT_1}{V_1} = \frac{1 \text{ mol} \times 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 100 \text{ K}}{1 \text{ L}} P1=8.21 atmP_1 = 8.21 \text{ atm}

2. Determine the constant (kk) from the given relation The given relation is P/V2=constant=kP/V^2 = \text{constant} = k. Using the initial conditions (P1,V1P_1, V_1): k=P1V12=8.21 atm(1 L)2=8.21 atm L−2k = \frac{P_1}{V_1^2} = \frac{8.21 \text{ atm}}{(1 \text{ L})^2} = 8.21 \text{ atm L}^{-2} So, the pressure can be expressed as a function of volume: P=kV2=8.21V2P = kV^2 = 8.21 V^2.

3. Calculate the work done (WW) For a reversible process, the work done is given by the integral: W=−∫V1V2PdVW = -\int_{V_1}^{V_2} P dV Substitute the expression for PP: W=−∫12(8.21V2)dVW = -\int_{1}^{2} (8.21 V^2) dV W=−8.21∫12V2dVW = -8.21 \int_{1}^{2} V^2 dV Integrate V2V^2: W=−8.21[V33]12W = -8.21 \left[ \frac{V^3}{3} \right]_{1}^{2} Now, apply the limits of integration (V2=2 LV_2 = 2 \text{ L}, V1=1 LV_1 = 1 \text{ L}): W=−8.21(233−133)W = -8.21 \left( \frac{2^3}{3} - \frac{1^3}{3} \right) W=−8.21(83−13)W = -8.21 \left( \frac{8}{3} - \frac{1}{3} \right) W=−8.21(73)W = -8.21 \left( \frac{7}{3} \right) W=−57.473 L atmW = -\frac{57.47}{3} \text{ L atm} W≈−19.1567 L atmW \approx -19.1567 \text{ L atm}

4. Convert the work done to calories Given the conversion factor: 1 L atm=24 cal1 \text{ L atm} = 24 \text{ cal}. W=−19.1567 L atm×24 cal/L atmW = -19.1567 \text{ L atm} \times 24 \text{ cal/L atm} W=−459.76 calW = -459.76 \text{ cal}

The negative sign indicates that work is done by the gas on the surroundings (i.e., the system loses energy as work).

The final answer is −459.76 cal\boxed{-459.76 \text{ cal}}.

Explanation of the solution:

  1. Used ideal gas law (P1V1=nRT1P_1V_1 = nRT_1) to find initial pressure P1=8.21 atmP_1 = 8.21 \text{ atm}.
  2. Used the given relation P/V2=constantP/V^2 = \text{constant} to find the constant k=P1/V12=8.21 atm/L2k = P_1/V_1^2 = 8.21 \text{ atm/L}^2, yielding P=8.21V2P = 8.21 V^2.
  3. Calculated work done using the integral W=−∫V1V2PdV=−∫128.21V2dVW = -\int_{V_1}^{V_2} P dV = -\int_{1}^{2} 8.21 V^2 dV.
  4. Evaluated the integral: W=−8.21[V33]12=−8.21(233−133)=−8.21(73)=−19.1567 L atmW = -8.21 \left[ \frac{V^3}{3} \right]_{1}^{2} = -8.21 \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = -8.21 \left( \frac{7}{3} \right) = -19.1567 \text{ L atm}.
  5. Converted the result from L atm to calories using 1 L atm=24 cal1 \text{ L atm} = 24 \text{ cal}: W=−19.1567×24=−459.76 calW = -19.1567 \times 24 = -459.76 \text{ cal}.