Internal energy of a gas is given as U = 3nRT. 1 mole of He gas takes 126 J of heat and its temperature rises by 4°C. Atmospheric pressure is \(P_0 = 10^5\) Pa and area of piston is 17 cm\(^2\). Find distance moved by piston.
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For any ideal gas, the work done \(W\) in an isobaric process is related to the heat supplied \(\Delta Q\) by \(W = \Delta Q \frac{R}{C_P}\). Finding this ratio can sometimes be a shortcut to calculating the work done.
Step 1: Understanding the Question:
A gas in a cylinder with a freely moving piston is heated. This is an isobaric (constant pressure) process, as the piston moves against the constant atmospheric pressure. We need to find how far the piston moves (\(\Delta x\)) using the First Law of Thermodynamics. Step 2: Key Formula or Approach:
1. First Law of Thermodynamics: \(\Delta Q = \Delta U + W\).
2. Internal Energy: From \(U = 3nRT\), we find the change \(\Delta U = 3nR\Delta T\). This also implies the molar heat capacity at constant volume is \(C_V = 3R\).
3. Work Done in Isobaric Process: \(W = P\Delta V = P_0 (A \cdot \Delta x)\). Also, for an ideal gas, \(W = nR\Delta T\).
4. Heat Supplied in Isobaric Process: \(\Delta Q = nC_P\Delta T\), where \(C_P = C_V + R\). Step 3: Detailed Explanation:
First, let's determine the properties of the gas based on its internal energy.
Given \(U = 3nRT\). The molar heat capacity at constant volume is \(C_V = \frac{1}{n} \frac{dU}{dT} = 3R\).
Since the process is isobaric (piston is free to move), the molar heat capacity at constant pressure is:
\[ C_P = C_V + R = 3R + R = 4R \]
Now, we can relate the heat supplied (\(\Delta Q\)) to the work done (\(W\)).
\[ \Delta Q = nC_P\Delta T = n(4R)\Delta T = 4(nR\Delta T) \]
The work done in an isobaric process is \(W = nR\Delta T\).
Comparing the two expressions, we see that:
\[ W = \frac{1}{4} \Delta Q \]
We are given \(\Delta Q = 126\) J. So, the work done by the gas is:
\[ W = \frac{1}{4} \times 126 \text{ J} = 31.5 \text{ J} \]
This work is done by expanding the gas against the atmospheric pressure, moving the piston.
\[ W = P_0 \Delta V = P_0 \cdot A \cdot \Delta x \]
We need to solve for the distance \(\Delta x\). Make sure all units are in SI.
- \(W = 31.5\) J.
- \(P_0 = 10^5\) Pa.
- Area \(A = 17 \text{ cm}^2 = 17 \times 10^{-4} \text{ m}^2\).
\[ 31.5 = (10^5) \times (17 \times 10^{-4}) \times \Delta x \]
\[ 31.5 = (17 \times 10^1) \times \Delta x = 170 \Delta x \]
\[ \Delta x = \frac{31.5}{170} \text{ m} \approx 0.1853 \text{ m} \]
Converting the distance to centimeters:
\[ \Delta x = 0.1853 \times 100 \text{ cm} = 18.53 \text{ cm} \]
Step 4: Final Answer:
The distance moved by the piston is approximately 18.5 cm.
