A gas at 155 kPa and 25'C has an initial volume of 1.00 L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?

Question

Peyton Adams · Accepted Answer

#V_2=0.342*L#...the gas is compressed....

We use the old #"combined gas law..."# #(P_1V_1)/T_1=(P_2V_2)/T_2# And if we solve for #V_2=(P_1V_1)/T_1xxT_2/P_2#...the which product CLEARLY has the units of #"volume"#. And we plug in the values.... #V_2=(155*cancel(kPa)xx1.00*Lxx398*cancelK)/(298*cancelKxx605*cancel(kPa))=0.342*L# Here, pressure prevails over temperature.

Wyatt Adkins · Answer

undefined To find the new volume of the gas, we can use the combined gas law, which states:

$ P_1 \times V_1 / T_1 = P_2 \times V_2 / T_2 $

Given:
$ P_1 = 155 $ kPa
$ V_1 = 1.00 $ L
$ T_1 = 25^\circ $C + 273.15 = 298.15 K
$ P_2 = 605 $ kPa
$ T_2 = 125^\circ $C + 273.15 = 398.15 K

We can rearrange the equation to solve for $ V_2 $:

$ V_2 = (P_1 \times V_1 \times T_2) / (P_2 \times T_1) $

Substitute the values:

$ V_2 = (155 \times 1.00 \times 398.15) / (605 \times 298.15) $

$ V_2 = (61651.7) / (180520.75) $

$ V_2 ≈ 0.34 $ L

Therefore, the new volume of the gas is approximately 0.34 L.