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A 7 L sample of gas has a pressure of 1.1 atm at a temperature of 285 K. If the pressure decreases to 0.6 atm, causing the volume to increase to 10 L, what is the new temperature? Round your answer to the nearest tenth.

Respuesta :

The answer is 222.1 K

Answer : The new temperature is, 222.1 K

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 1.1 atm

[tex]P_2[/tex] = final pressure of gas = 0.6 atm

[tex]V_1[/tex] = initial volume of gas = 7 L

[tex]V_2[/tex] = final volume of gas = 10 L

[tex]T_1[/tex] = initial temperature of gas = 285 K

[tex]T_2[/tex] = final temperature of gas = ?

Now put all the given values in the above equation, we get:

[tex]\frac{1.1atm\times 7L}{285K}=\frac{0.6atm\times 10L}{T_2}[/tex]

[tex]T_2=222.07K\approx 222.1K[/tex]

Therefore, the new temperature is, 222.1 K

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