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Two moles of an ideal monoatomic gas at ...

Two moles of an ideal monoatomic gas at `27^(@)C`occupies a volume of V . If the gas is expanded adiabatically to the volume , 2V then the work done by the gas will be `[lambda=5//3,R=8.31j//molK]`

A

`-2767.23j`

B

`2767.23j`

C

`2500j`

D

`-2500j`

Text Solution

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The correct Answer is:
To solve the problem of finding the work done by an ideal monoatomic gas during an adiabatic expansion from volume \( V \) to \( 2V \), we can follow these steps: ### Step 1: Identify the Given Information - Number of moles, \( n = 2 \) - Initial temperature, \( T_1 = 27^\circ C = 300 \, K \) (after converting to Kelvin) - Initial volume, \( V_1 = V \) - Final volume, \( V_2 = 2V \) - Heat capacity ratio, \( \gamma = \frac{5}{3} \) - Ideal gas constant, \( R = 8.31 \, J/(mol \cdot K) \) ### Step 2: Use the Adiabatic Condition For an adiabatic process, we have the relation: \[ T V^{\gamma - 1} = \text{constant} \] This implies: \[ \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma - 1} \] Substituting the values: \[ \frac{T_1}{T_2} = \left(\frac{2V}{V}\right)^{\frac{5}{3} - 1} = 2^{\frac{2}{3}} \] ### Step 3: Solve for \( T_2 \) Rearranging gives: \[ T_2 = T_1 \cdot 2^{-\frac{2}{3}} \] Substituting \( T_1 = 300 \, K \): \[ T_2 = 300 \cdot 2^{-\frac{2}{3}} \approx 300 \cdot 0.63 \approx 189 \, K \] ### Step 4: Calculate Change in Internal Energy The change in internal energy \( \Delta U \) for a monoatomic ideal gas is given by: \[ \Delta U = n C_v \Delta T \] Where \( C_v = \frac{3}{2} R \). Thus: \[ \Delta U = n \cdot \frac{3}{2} R (T_2 - T_1) \] Substituting the values: \[ \Delta U = 2 \cdot \frac{3}{2} \cdot 8.31 \cdot (189 - 300) \] Calculating \( (189 - 300) = -111 \): \[ \Delta U = 2 \cdot \frac{3}{2} \cdot 8.31 \cdot (-111) \] \[ \Delta U = -2 \cdot 3 \cdot 8.31 \cdot 111 \approx -5533.26 \, J \] ### Step 5: Calculate Work Done by the Gas For an adiabatic process, the work done \( W \) is related to the change in internal energy: \[ W = -\Delta U \] Thus: \[ W = 5533.26 \, J \] ### Final Answer The work done by the gas during the adiabatic expansion is approximately: \[ W \approx 5533.26 \, J \] ---
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