A $780 \cdot m L$ $780mL$ volume of gas $1.00 \cdot a t m$ $1.00atm$ pressure, and $310 \cdot K$ $310K$ temperature, is cooled to $295 \cdot K$ $295K$ , and $65 \cdot m m \cdot H g$ $65mmHg$ pressure. What is the new volume?

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A $780 \cdot m L$ $780mL$ volume of gas $1.00 \cdot a t m$ $1.00atm$ pressure, and $310 \cdot K$ $310K$ temperature, is cooled to $295 \cdot K$ $295K$ , and $65 \cdot m m \cdot H g$ $65mmHg$ pressure. What is the new volume?

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Answer 1 · 2016-11-14T15:40:28

For a given quantity of gas, ......... $V_{2} = 8.70 \cdot L .$ $V_2=8.70*L.$

Explanation:

For a given quantity of gas, the combined gas law holds that $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ , temperature quoted in $degrees Kelvin$ $"degrees Kelvin"$ .

Thus $V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}$ $V_2=(P_1V_1)/T_1xxT_2/P_2$

$= \frac{1.00 \cdot a t m \times 780 \cdot m L}{310 \cdot K} \times \frac{295 \cdot K}{\frac{65 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}}}$ $=(1.00*atmxx780*mL)/(310*K)xx(295*K)/((65*mm*Hg)/(760*mm*Hg*atm^-1)$

$= 8.69 \times 10^{3} \cdot m L = 8.70 \cdot L$ $=8.69xx10^3*mL=8.70*L$

We use here the useful relationship that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$ $760*mm*Hg-=1*atm$ .

And thus a measurement of length on a mercury manometer, which is fairly easy to do in a lab, can be used to solve Gas law problems.

Please note that I used $22$ $22$ $^{\circ} C$ $""^@C$ , $\equiv$ $-=$ $295 \cdot K$ $295*K$ , and $37$ $37$ $^{\circ} C$ $""^@C$ , $\equiv$ $-=$ $310 \cdot K$ $310*K$ . You can change the calculation accordingly if I have got your starting conditions wrong.

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A $780 \cdot m L$ $780mL$ volume of gas $1.00 \cdot a t m$ $1.00atm$ pressure, and $310 \cdot K$ $310K$ temperature, is cooled to $295 \cdot K$ $295K$ , and $65 \cdot m m \cdot H g$ $65mmHg$ pressure. What is the new volume?

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Explanation:

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A 780⋅mL780*mL volume of gas 1.00⋅atm1.00*atm pressure, and 310⋅K310*K temperature, is cooled to 295⋅K295*K, and 65⋅mm⋅Hg65*mm*Hg pressure. What is the new volume?

1 Answer

Explanation:

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A $780 \cdot m L$ $780mL$ volume of gas $1.00 \cdot a t m$ $1.00atm$ pressure, and $310 \cdot K$ $310K$ temperature, is cooled to $295 \cdot K$ $295K$ , and $65 \cdot m m \cdot H g$ $65mmHg$ pressure. What is the new volume?