hat is the final temperature of the gas if the 13. A sample of a monatomice ideal gas is originally at 20 pressure is doubled and volume is reduced to one-fourth its initial value? A) 900 °C B) 10 °C C) 80 °C D) -130 °C E) -260 °C
hat is the final temperature of the gas if the 13. A sample of a monatomice ideal gas is originally at 20 pressure is doubled and volume is reduced to one-fourth its initial value? A) 900 °C B) 10 °C C) 80 °C D) -130 °C E) -260 °C
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![**Question 13:**
A sample of a monatomic ideal gas is originally at 20°C. What is the final temperature of the gas if the pressure is doubled and the volume is reduced to one-fourth its initial value?
**Options:**
A) 900°C
B) 10°C
C) 80°C
D) –130°C
E) –260°C
*Explanation:*
This question deals with gas laws. Assuming the sample behaves as an ideal gas, you can use the ideal gas law equation in its combined form, considering the initial and final states of the gas:
\[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \]
Where:
- \(P_1\), \(V_1\), and \(T_1\) are the initial pressure, volume, and temperature.
- \(P_2\), \(V_2\), and \(T_2\) are the final pressure, volume, and temperature.
Given:
- Initial temperature (\(T_1\)) = 293 K (converting 20°C to Kelvin)
- Final pressure is double the initial pressure (\(P_2 = 2P_1\))
- Final volume is one-fourth the initial volume (\(V_2 = V_1/4\))
Plug these into the formula to solve for \(T_2\) to find the final temperature.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0fd57332-4152-4982-80d5-c019d57ce2f7%2F02b8183b-e6ea-42aa-9511-1ba23f0c2d93%2Ftztiejap_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 13:**
A sample of a monatomic ideal gas is originally at 20°C. What is the final temperature of the gas if the pressure is doubled and the volume is reduced to one-fourth its initial value?
**Options:**
A) 900°C
B) 10°C
C) 80°C
D) –130°C
E) –260°C
*Explanation:*
This question deals with gas laws. Assuming the sample behaves as an ideal gas, you can use the ideal gas law equation in its combined form, considering the initial and final states of the gas:
\[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \]
Where:
- \(P_1\), \(V_1\), and \(T_1\) are the initial pressure, volume, and temperature.
- \(P_2\), \(V_2\), and \(T_2\) are the final pressure, volume, and temperature.
Given:
- Initial temperature (\(T_1\)) = 293 K (converting 20°C to Kelvin)
- Final pressure is double the initial pressure (\(P_2 = 2P_1\))
- Final volume is one-fourth the initial volume (\(V_2 = V_1/4\))
Plug these into the formula to solve for \(T_2\) to find the final temperature.
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