Question

a helium balloon has a volume of 10.2 l at -34°c and 366 mm hg. what is its volume at stp?

4.77 l
6.12 l
5.61 l
18.5 l
4.37 l

Explanation:

Step1: Convert temperature to Kelvin

The initial temperature \(T_1=-34^{\circ}C\). Convert to Kelvin: \(T_1 = - 34+273 = 239K\). At STP, \(T_2 = 273K\) and \(P_2=760mmHg\). The initial volume \(V_1 = 10.2L\) and initial pressure \(P_1 = 366mmHg\).

Step2: Use the combined gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

Rearrange for \(V_2\): \(V_2=\frac{P_1V_1T_2}{P_2T_1}\)
Substitute the values: \(V_2=\frac{366\times10.2\times273}{760\times239}\)
First, calculate the numerator: \(366\times10.2\times273=(366\times10.2)\times273 = 3733.2\times273=1019163.6\)
Then, calculate the denominator: \(760\times239 = 181640\)
Now, \(V_2=\frac{1019163.6}{181640}\approx5.61L\)

Answer:

5.61 L