For a fixed amount of gas,the volume is x L | vedclass

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(A) According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given $V_1 = V_2 = x \ L$,the equation simplifies to $\frac{P_1}{T_1} =

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$2y \ atm$ and $2z \ K$

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(A) According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given $V_1 = V_2 = x \ L$,the equation simplifies to $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.Substituting the initial values,we get $\frac{y}{z} = \frac{P_2}{T_2}$.For the volume to remain $x \ L$,the ratio of pressure to temperature must remain constant,i.e.,$\frac{P_2}{T_2} = \frac{y}{z}$.Checking option $A$: $\frac{2y}{2z} = \frac{y}{z}$.Thus,at $2y \ atm$ and $2z \ K$,the volume remains $x \ L$.

For a fixed amount of gas,the volume is $x \ L$ at a pressure of $y \ atm$ and a temperature of $z \ K$. At which of the following conditions will the volume remain $x \ L$? (Consider $y \neq z$)

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