At constant temperature, a $3.8 \cdot L$ $3.8L$ volume of gas at $765 \cdot m m \cdot H g$ $765mmHg$ , was expanded to give a pressure of $0.500 \cdot a t m$ $0.500atm$ . What is the new volume?

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At constant temperature, a $3.8 \cdot L$ $3.8L$ volume of gas at $765 \cdot m m \cdot H g$ $765mmHg$ , was expanded to give a pressure of $0.500 \cdot a t m$ $0.500atm$ . What is the new volume?

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Answer 1 · 2017-05-04T06:14:54

$P_{1} V_{1} = P_{2} V_{2}$ $P_1V_1=P_2V_2$ under conditions of constant temperature.

Explanation:

We need to know that $1 \cdot a t m$ $1*atm$ will support a column of mercury that is $760 \cdot m m$ $760*mm$ high, and thus a mercury column may used as a highly visual measurement of pressure:

$1 \cdot a t m \equiv 760 \cdot m m \cdot H g$ $1*atm-=760*mm*Hg$ (note that given the schemozzle that occurs when you spill mercury in the lab, you really should not put mercury under a pressure of MORE than 1 atmosphere). Here.......

$P_{1} = \frac{765 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}} = 1.01 \cdot a t m .$ $P_1=(765*mm*Hg)/(760*mm*Hg*atm^-1)=1.01*atm.$

We solve for $V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{1.01 \cdot a t m \times 3.8 \cdot L}{0.500 \cdot a t m} = 7.60 \cdot L$ $V_2=(P_1V_1)/P_2=(1.01*atmxx3.8*L)/(0.500*atm)=7.60*L$

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At constant temperature, a $3.8 \cdot L$ $3.8L$ volume of gas at $765 \cdot m m \cdot H g$ $765mmHg$ , was expanded to give a pressure of $0.500 \cdot a t m$ $0.500atm$ . What is the new volume?

1 Answer

Explanation:

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Humanities

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At constant temperature, a 3.8⋅L3.8*L volume of gas at 765⋅mm⋅Hg765*mm*Hg, was expanded to give a pressure of 0.500⋅atm0.500*atm. What is the new volume?

1 Answer

Explanation:

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At constant temperature, a $3.8 \cdot L$ $3.8L$ volume of gas at $765 \cdot m m \cdot H g$ $765mmHg$ , was expanded to give a pressure of $0.500 \cdot a t m$ $0.500atm$ . What is the new volume?