Question:

cylinder
Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18C18^\circ \text{C}. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm\text{L atm} work is done in this reversible process. x=x = ______ L atm\text{L atm}. (nearest integer)
[Given: Absolute temperature=C+273.15,R=0.08206L atm mol1K1][ \text{Given: Absolute temperature} = ^\circ \text{C} + 273.15, \, R = 0.08206 \, \text{L atm mol}^{-1} \text{K}^{-1} ]

Updated On: Mar 21, 2025
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Correct Answer: 55

Solution and Explanation

Work done (WW) in an isothermal reversible expansion of an ideal gas is given by:
W=nRTln(V2V1)W = -nRT \ln \left( \frac{V_2}{V_1} \right)
Given:
n=1mol,T=18C=18+273.15=291.15Kn = 1 \, \text{mol}, \quad T = 18^\circ \text{C} = 18 + 273.15 = 291.15 \, \text{K}
V1=10L,V2=100LV_1 = 10 \, \text{L}, \quad V_2 = 100 \, \text{L}
Substitute the values:
W=1×0.08206×291.15×ln(10010)W = -1 \times 0.08206 \times 291.15 \times \ln \left( \frac{100}{10} \right)
W1×0.08206×291.15×ln(10)W \approx -1 \times 0.08206 \times 291.15 \times \ln(10)
W55.0128L atmW \approx -55.0128 \, \text{L atm}
The work done by the system is approximately 55L atm-55 \, \text{L atm} (rounded to nearest integer).

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