Question

1. a given sample of gas has a volume of 4.20 l at 60.0 °c and 1.00 atm pressure. calculate its pressure if the volume is changed to 5.00 l and the temperature to 27.0 °c.

v1 = 4.2 v2 = 5
t1 = 333k t2 = 300
p1 = 1.0 p2 =?

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 60.0 + 273 = 333\ K$, $T_2=27.0 + 273 = 300\ K$

Step2: Apply the combined - gas law formula

The combined - gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. We know that $P_1 = 1.00\ atm$, $V_1 = 4.20\ L$, $V_2 = 5.00\ L$, $T_1 = 333\ K$, and $T_2 = 300\ K$. Rearranging the formula for $P_2$ gives $P_2=\frac{P_1V_1T_2}{V_2T_1}$.

Step3: Substitute the values and calculate

$P_2=\frac{1.00\ atm\times4.20\ L\times300\ K}{5.00\ L\times333\ K}=\frac{1260}{1665}\ atm\approx0.757\ atm$

Answer:

$0.757\ atm$