Answer:
288K
Explanation:
In most situation where the name of the gas is not mentioned, the best option we have is to use the ideal gas law.
Remember, the pressure was doubled, and the volume was halved, numerically cancelling each other out. As a result, the final temperature be the same as the initial temperature.
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{(1 atm)(3L)}{(0.08257L.atm/mol.k)(288k)}[/tex]
[tex]n= 0.1269 mol gas[/tex]
Now, solving for T
[tex]T = \frac{PV}{nR}[/tex]
[tex]T=\frac{(2 atm)(1.5L)}{(0.1269mol gas) (0.082057 L.atm/mol.K)}[/tex]
[tex]T=288K[/tex]
The temperature of the sample will be "288 K".
According to the question,
Pressure,
Volume,
Temperature,
By using the Ideal gas equation, we get
→ [tex]\frac{P_1 V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
By substituting the values, we get
→ [tex]\frac{1\times 3}{288} = \frac{2\times 1.5}{T_2}[/tex]
→ [tex]\frac{3}{288} = \frac{3}{T_2}[/tex]
→ [tex]T_2 = 288 \ K[/tex]
Thus the above answer is right.
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