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

15964 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-21T03:04:45

$0.34 L$ $"0.34 L"$

Ideal gas equation is

$P V = n R T$ $PV = nRT$

$\frac{P V}{T} = n R$ $(PV)/T = nR$

In both the cases $\frac{P V}{T}$ $(PV)/T$ is equal to $n R$ $nR$

Therefore,

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

$V_{2} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}$ $V_2 = (P_1V_1)/T_1 × T_2/P_2$

$V_2 = (155 cancel"kPa" × "1.00 L")/(298 cancel"K") × (398 cancel"K")/(605 cancel"kPa") = "0.34 L"$