Question #d9f89

Question

Question #d9f89

1 Answer

Impact of this question

2320 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-29T21:02:37

The new volume wil be 3 L.

Because only the number of moles of gas is kept constant, you're going to have to use the combined gas law to solve for the new volume of the gas sample.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$ , where

$P_{1}$ $P_1$ , $V_{1}$ $V_1$ , $T_{1}$ $T_1$ - the pressure, volume, and temperature at an initial state;
$P_{2}$ $P_2$ , $V_{2}$ $V_2$ , $T_{2}$ $T_2$ - the pressure, volume, and temperature at a final state.

Rearrange the above equation to isolate $V_{2}$ $V_2$

$V_{2} = \frac{P_{1}}{P_{2}} \cdot \frac{T_{2}}{T_{1}} \cdot V_{1}$ $V_2 = P_1/P_2 * T_2/T_1 * V_1$

Make sure that you use the same units for pressure and that you convert the temperature from degrees Celsius to Kelvin

$V_2 = (1.00cancel("atm"))/(150/760cancel("atm")) * ((273.15)cancel("K"))/((273.15 + 36)cancel("K")) * "0.600 L"$

$V_2 = "2.685 L"$

Rounded to one sig fig, the number of sig figs given for $0^@"C"$ , the answer will be

$V_2 = color(green)("3 L")$

Science

Math

Humanities

... and beyond

Question #d9f89

1 Answer

Related questions

Impact of this question