If the volume of a gas is $5.9 \cdot m L$ $5.9mL$ when $T = 279.05 \cdot K$ $T=279.05K$ and $P = 705.0 \cdot m m \cdot H g$ $P=705.0mmHg$ , how will volume change when the gas is warmed to $332.8 \cdot K$ $332.8K$ , and the pressure is increased to $761.1 \cdot m m \cdot H g$ $761.1mm*Hg$ ?

Question

If the volume of a gas is $5.9 \cdot m L$ $5.9mL$ when $T = 279.05 \cdot K$ $T=279.05K$ and $P = 705.0 \cdot m m \cdot H g$ $P=705.0mmHg$ , how will volume change when the gas is warmed to $332.8 \cdot K$ $332.8K$ , and the pressure is increased to $761.1 \cdot m m \cdot H g$ $761.1mm*Hg$ ?

1 Answer

Impact of this question

2162 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-24T05:27:23

The key to answering this question is the knowledge that $1 \cdot a t m$ $1*atm$ of pressure will support a mercury column that is $760 \cdot m m$ $760*mm$ high.

Explanation:

A mercury column, a so-called mercury manometer, is used in many laboratories as a visual representation of how good your vacuum pump is; if you isolate your gas line from the pump, you can check to see if there are any leaks in your gas line, because the level of mercury will drop (quickly or slowly) with a pinhole leak.

You have probably also seen doctors use a sphygmomanometer, to check a patient's blood pressure.

![wisegeek.com]( useruploads.socratic.org )

Mercury is in fact being phased out of many laboratories (and probably also doctors' offices) in that it poses a safety concern, and if you break the glass (not an uncommon occurrence), you get mercury EVERYWHERE, and this is major cleanup job that contract cleaners will not touch. Anyway that is the background I wanted to include because it might seem a bit whack to use a unit of length to represent pressure.

We use the old Combined Gas Law, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ , and solve for $V_{2}$ $V_2$ . Temperature must be $tempertura assoluta$ $"tempertura assoluta"$ , where $0 \cdot K \equiv - 273.15$ $0*K-=-273.15$ $^{\circ} C$ $""^@C$ .

And so.......... $V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}} = \frac{\frac{705 \cdot m m \cdot H g \times 5.9 \cdot m L}{760 \cdot m m \cdot H g \cdot a t m^{- 1}}}{279.05 \cdot K} \times \frac{332.8 \cdot K}{\frac{761.1 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t m^{- 1}}}$ $V_2=(P_1V_1)/T_1xxT_2/P_2=((705*mm*Hgxx5.9*mL)/(760*mm*Hg*atm^-1))/(279.05*K)xx(332.8*K)/((761.1*mm*Hg)/(760*mm*Hg*atm^-1))$

I gets about $7 \cdot m L$ $7*mL$ , we are increasing the temperature $(volume ↑ ⏐)$ $("volume"uarr)$ , but ALSO INCREASING the pressure $(volume ⏐ ↓)$ $("volume"darr)$ . And if you think this is a lot of work for a little problem, I completely agree (of course I do, I solved it!). Look in your text and read about Boyle's Law, Charles' Law, and the Combined gas law. And ALSO you must learn that $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$ $1*atm-=760*mm*Hg$ .

Science

Math

Humanities

... and beyond

If the volume of a gas is $5.9 \cdot m L$ $5.9mL$ when $T = 279.05 \cdot K$ $T=279.05K$ and $P = 705.0 \cdot m m \cdot H g$ $P=705.0mmHg$ , how will volume change when the gas is warmed to $332.8 \cdot K$ $332.8K$ , and the pressure is increased to $761.1 \cdot m m \cdot H g$ $761.1mm*Hg$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If the volume of a gas is 5.9⋅mL5.9*mL when T=279.05⋅KT=279.05*K and P=705.0⋅mm⋅HgP=705.0*mm*Hg, how will volume change when the gas is warmed to 332.8⋅K332.8*K, and the pressure is increased to 761.1⋅mm⋅Hg761.1*mm*Hg?

1 Answer

Explanation:

Related questions

Impact of this question

If the volume of a gas is $5.9 \cdot m L$ $5.9mL$ when $T = 279.05 \cdot K$ $T=279.05K$ and $P = 705.0 \cdot m m \cdot H g$ $P=705.0mmHg$ , how will volume change when the gas is warmed to $332.8 \cdot K$ $332.8K$ , and the pressure is increased to $761.1 \cdot m m \cdot H g$ $761.1mm*Hg$ ?