A gas takes up a volume of 19.5 liters, has a pressure of 2.9 atm, and a temperature of 5.5°C. If I raise the temperature to 65°C and lower the pressure to 1.5 atm, what is the new volume of the gas?

Question

2033 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-01T19:33:37

$V_{2} = 45.8 L$ $V_2=45.8L$

From the ideal gas law $P V = n R T$ $PV=nRT$ we can conclude that $n = \frac{P V}{R T}$ $n=(PV)/(RT)$ .

If the pressure $P$ $P$ , the volume $V$ $V$ and the temperature $T$ $T$ of the gas change between two points, this change can be illustrated by:

$n = \frac{P_{1} V_{1}}{R T_{1}} = \frac{P_{2} V_{2}}{R T_{2}}$ $n=(P_1V_1)/(RT_1)=(P_2V_2)/(RT_2)$

Therefore, this expression can be modified as:

$(P_1V_1)/(cancel(R)T_1)=(P_2V_2)/(cancel(R)T_2)=>(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$

Thus, $V_2=(P_1V_1)/(T_1)xx(T_2)/(P_2)$

$V_2=(2.9cancel(atm)xx19.5L)/(278.5cancel(K))xx(338cancel(K))/(1.5cancel(atm))=45.8L$