Question:
At constant temperature, one mole of an ideal gas of volume 2L expanded to 100 L against an external pressure of 1 atm under reversible conditions. What is the work done (in J)? (1 L atm = 101.3 J; \(\log 5 = 0.7\))
At constant temperature, one mole of an ideal gas of volume 2L expanded to 100 L against an external pressure of 1 atm under reversible conditions. What is the work done (in J)? (1 L atm = 101.3 J; \(\log 5 = 0.7\))
Show Hint
Expansion work is negative by convention because the system loses energy to the surroundings.
Updated On: Jun 9, 2026
- -163.3
- -793.2
- +793.2
- +326.6
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
For a reversible isothermal expansion of an ideal gas, the work done is given by the integral of pressure with respect to volume.
Step 1: Apply the reversible work formula.
The formula for reversible isothermal work is: \[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \] Since \(PV = nRT\), we can substitute \(nRT = P_1V_1\). Assuming expansion from 1 atm at 2L: \[ W = -P_1V_1 \ln\left(\frac{V_2}{V_1}\right) \] \[ W = -(1 \text{ atm} \times 2 \text{ L}) \times 2.303 \log\left(\frac{100}{2}\right) \]
Step 2: Calculate the numerical value.
\[ W = -2 \times 2.303 \times \log(50) \] Given \(\log 5 = 0.7\), then \(\log 50 = 1 + 0.7 = 1.7\). \[ W = -2 \times 2.303 \times 1.7 = -7.83 \text{ L atm} \]
Step 3: Convert to Joules.
\[ W = -7.83 \times 101.3 \text{ J} \approx -793.2 \text{ J} \] \[ \boxed{-793.2} \]
Step 1: Apply the reversible work formula.
The formula for reversible isothermal work is: \[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \] Since \(PV = nRT\), we can substitute \(nRT = P_1V_1\). Assuming expansion from 1 atm at 2L: \[ W = -P_1V_1 \ln\left(\frac{V_2}{V_1}\right) \] \[ W = -(1 \text{ atm} \times 2 \text{ L}) \times 2.303 \log\left(\frac{100}{2}\right) \]
Step 2: Calculate the numerical value.
\[ W = -2 \times 2.303 \times \log(50) \] Given \(\log 5 = 0.7\), then \(\log 50 = 1 + 0.7 = 1.7\). \[ W = -2 \times 2.303 \times 1.7 = -7.83 \text{ L atm} \]
Step 3: Convert to Joules.
\[ W = -7.83 \times 101.3 \text{ J} \approx -793.2 \text{ J} \] \[ \boxed{-793.2} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Thermodynamics Questions
- For one mole of an ideal gas, three isochors were obtained at \(V_1, V_2\) and \(V_3\) respectively. Their slopes are \(m_1, m_2\) and \(m_3\). If \(V_1<V_2<V_3\), then the correct relationship of slopes is:
- The enthalpy change (ΔH) for the reaction N2(g) + 3H2(g) → 2NH3(g) is -92 kJ/mol.
What is the enthalpy change for the decomposition of 1 mol of NH3(g) into its elements? - Identify the correct statements from the following. (i). For adiabatic process, $\Delta U = W_{adiabatic$} (ii). Work is a path function. (iii). Volume is an extensive property.
- Hydration enthalpy of which of the following ions is highest?
- The standard Gibbs energy ($\Delta G^\ominus$) for the following reaction is A(s) + B$^{2+}$ (aq) $\rightleftharpoons$ A$^{2+}$ (aq) + B(s), K$_c$=10$^{12}$ at 25$^\circ$C (K$_c$= equilibrium constant)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions