An ideal gas expands at a constant pressure of 7.0 atm from 280 mL to 630 mL. Heat then flows out of the gas, at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate the total heat flow into the gas.

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript and apply the relevant thermodynamic principles.

### Step 1: Identify the Initial and Final Conditions
- The gas expands from an initial volume \( V_1 = 280 \, \text{mL} \) to a final volume \( V_2 = 630 \, \text{mL} \) at a constant pressure of \( P = 7.0 \, \text{atm} \).

### Step 2: Convert Units
- Convert the pressure from atm to Pascal:
\[ P = 7.0 \, \text{atm} \times 1.01325 \times 10^5 \, \text{Pa/atm} = 7.0 \times 1.01325 \times 10^5 \, \text{Pa} \approx 7.0 \times 101325 \, \text{Pa} = 710325 \, \text{Pa} \]
- Convert the volumes from mL to cubic meters:
\[ V_1 = 280 \, \text{mL} = 280 \times 10^{-6} \, \text{m}^3 = 2.8 \times 10^{-4} \, \text{m}^3 \]
\[ V_2 = 630 \, \text{mL} = 630 \times 10^{-6} \, \text{m}^3 = 6.3 \times 10^{-4} \, \text{m}^3 \]

### Step 3: Calculate the Change in Volume
- The change in volume \( \Delta V \) is:
\[ \Delta V = V_2 - V_1 = (6.3 \times 10^{-4} - 2.8 \times 10^{-4}) \, \text{m}^3 = 3.5 \times 10^{-4} \, \text{m}^3 \]

### Step 4: Calculate the Work Done by the Gas
- The work done \( W \) during the expansion at constant pressure is given by:
\[ W = P \Delta V \]
Substituting the values:
\[ W = 710325 \, \text{Pa} \times 3.5 \times 10^{-4} \, \text{m}^3 \approx 248.3 \, \text{J} \]

### Step 5: Determine the Heat Flow
- Since the gas is allowed to cool at constant volume back to its original temperature, the heat flow \( Q \) can be determined using the first law of thermodynamics:
\[ Q = \Delta U + W \]
- However, since the temperature returns to its original value, the change in internal energy \( \Delta U \) is zero (for an ideal gas, internal energy depends only on temperature). Therefore:
\[ Q = 0 + W = W \]
- Thus, the total heat flow into the gas is:
\[ Q = 248.3 \, \text{J} \]

### Final Answer
The total heat flow into the gas is approximately **248 J**.

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