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a gas has a volume of 2.00 at 323k and 3.00 ATM What would be the new volume if the temperature is changed to 273 K in the pressure is changed to 1ATM

Sagot :

Cromulentidn

Answer:

[tex]V_2 = 5.07L[/tex]

Explanation:

[tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\\V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{3.00atm \cdot 2.00L \cdot 273K}{323K \cdot 1.00atm} = 5.07L[/tex]

Apply combine gas law

[tex]\\ \rm\Rrightarrow \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{3(2)}{323}=\dfrav{V_2}{273}[/tex]

[tex]\\ \rm\Rrightarrow 6(273)=323V_2[/tex]

[tex]\\ \rm\Rrightarrow 1638=323V_2[/tex]

[tex]\\ \rm\Rrightarrow V_2=5.07L[/tex]

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