Question:
One mole of a monatomic ideal gas undergoes a transformation from an initial state with temperature $290\text{ K}$ and volume $30\text{ litres}$ to a final state with temperature $310\text{ K}$ and volume $16\text{ litres}$. On the pressure–volume ($P - V$) diagram, this process is represented by a straight line path. The magnitude of the work done (in joules) during this process is close to
One mole of a monatomic ideal gas undergoes a transformation from an initial state with temperature $290\text{ K}$ and volume $30\text{ litres}$ to a final state with temperature $310\text{ K}$ and volume $16\text{ litres}$. On the pressure–volume ($P - V$) diagram, this process is represented by a straight line path. The magnitude of the work done (in joules) during this process is close to
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Remember to convert volume from litres to $\text{m}^3$ ($1\text{ L} = 10^{-3}\text{ m}^3$) to keep all units in the standard SI system.
Updated On: Jun 16, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem describes an ideal gas undergoing a thermodynamic process along a straight line on a $P-V$ diagram.
We need to compute the work done, which corresponds to the area under this straight line.
Step 2: Key Formula or Approach:
- Use the ideal gas law to calculate the pressures at the initial and final states:
\[ P = \frac{nRT}{V} \]
- The work done $W$ along a straight-line path on a $P-V$ diagram is the area of the trapezoid:
\[ W = \frac{P_i + P_f}{2} (V_f - V_i) \]
Step 3: Detailed Explanation:
• The given values are $n = 1\text{ mole}$ and $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$.
• Initial state:
- $T_i = 290\text{ K}$
- $V_i = 30\text{ litres} = 30 \times 10^{-3}\text{ m}^3$
\[ P_i = \frac{nRT_i}{V_i} = \frac{1 \times 8.314 \times 290}{30 \times 10^{-3}} \approx 80369\text{ Pa} \]
• Final state:
- $T_f = 310\text{ K}$
- $V_f = 16\text{ litres} = 16 \times 10^{-3}\text{ m}^3$
\[ P_f = \frac{nRT_f}{V_f} = \frac{1 \times 8.314 \times 310}{16 \times 10^{-3}} \approx 161084\text{ Pa} \]
• Since the path is a straight line, the magnitude of the work done is:
\[ |W| = \frac{P_i + P_f}{2} |V_f - V_i| \]
\[ |W| = \frac{80369 + 161084}{2} \times (30 - 16) \times 10^{-3}\text{ m}^3 \]
\[ |W| = \frac{241453}{2} \times 14 \times 10^{-3} \]
\[ |W| = 241453 \times 7 \times 10^{-3} \approx 1690.17\text{ J} \]
Step 4: Final Answer:
The magnitude of the work done during this process is close to 1690 J.
The problem describes an ideal gas undergoing a thermodynamic process along a straight line on a $P-V$ diagram.
We need to compute the work done, which corresponds to the area under this straight line.
Step 2: Key Formula or Approach:
- Use the ideal gas law to calculate the pressures at the initial and final states:
\[ P = \frac{nRT}{V} \]
- The work done $W$ along a straight-line path on a $P-V$ diagram is the area of the trapezoid:
\[ W = \frac{P_i + P_f}{2} (V_f - V_i) \]
Step 3: Detailed Explanation:
• The given values are $n = 1\text{ mole}$ and $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$.
• Initial state:
- $T_i = 290\text{ K}$
- $V_i = 30\text{ litres} = 30 \times 10^{-3}\text{ m}^3$
\[ P_i = \frac{nRT_i}{V_i} = \frac{1 \times 8.314 \times 290}{30 \times 10^{-3}} \approx 80369\text{ Pa} \]
• Final state:
- $T_f = 310\text{ K}$
- $V_f = 16\text{ litres} = 16 \times 10^{-3}\text{ m}^3$
\[ P_f = \frac{nRT_f}{V_f} = \frac{1 \times 8.314 \times 310}{16 \times 10^{-3}} \approx 161084\text{ Pa} \]
• Since the path is a straight line, the magnitude of the work done is:
\[ |W| = \frac{P_i + P_f}{2} |V_f - V_i| \]
\[ |W| = \frac{80369 + 161084}{2} \times (30 - 16) \times 10^{-3}\text{ m}^3 \]
\[ |W| = \frac{241453}{2} \times 14 \times 10^{-3} \]
\[ |W| = 241453 \times 7 \times 10^{-3} \approx 1690.17\text{ J} \]
Step 4: Final Answer:
The magnitude of the work done during this process is close to 1690 J.
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