A quantity of gas occupies a volume of 804cm³ at a temperature of 127degree c . At what temperature will the volume of the gas be 603cm³ assuming that there is no change in the pressure
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Convert the initial temperature from Celsius to Kelvin.
To convert Celsius to Kelvin, add 273.15.
$$T_1 = 127 \, ^\circ\text{C} + 273.15 = 400.15 \, \text{K}$$
Step 2: Identify the given values.
Initial Volume ($V_1$) = $804 \, \text{cm}^3$
Initial Temperature ($T_1$) = $400.15 \, \text{K}$
Final Volume ($V_2$) = $603 \, \text{cm}^3$
Final Temperature ($T_2$) = ?
Step 3: Apply Charles's Law.
Charles's Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature:
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
To find $T_2$, rearrange the formula:
$$T_2 = \frac{V_2 T_1}{V_1}$$
Step 4: Substitute the values and calculate $T_2$.
$$T_2 = \frac{(603 \, \text{cm}^3) \times (400.15 \, \text{K})}{804 \, \text{cm}^3}$$
$$T_2 = \frac{241290.45 \, \text{K} \cdot \text{cm}^3}{804 \, \text{cm}^3}$$
$$T_2 \approx 300.11 \, \text{K}$$
Step 5: Convert the final temperature back to Celsius.
$$T_2 \, (^\circ\text{C}) = T_2 \, (\text{K}) - 273.15$$
$$T_2 \, (^\circ\text{C}) = 300.11 \, \text{K} - 273.15$$
$$T_2 \, (^\circ\text{C}) \approx 26.96 \, ^\circ\text{C}$$
The temperature at which the volume of the gas will be $603 \, \text{cm}^3$ is approximately $27.0 \, ^\circ\text{C}$.
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