Question

use the references to access important values if needed for this question.
(a) a sample of methane gas at a pressure of 0.929 atm and a temperature of 25.0 °c, occupies a volume of 807 ml. if the gas is allowed to expand at constant temperature until its pressure is 0.634 atm, the volume of the gas sample will be
(b) a sample of helium gas at a pressure of 0.859 atm and a temperature of 27.1 °c, occupies a volume of 16.3 liters. if the gas is allowed to expand at constant temperature to a volume of 27.3 liters, the pressure of the gas sample will be atm.
3 item attempts remaining

Explanation:

Step1: Convert temperature to Kelvin

For part (A):
The initial temperature \(T_1=(25.0 + 273.15)\text{K}=298.15\text{K}\)
For part (B):
The initial temperature \(T_1=(27.1+ 273.15)\text{K}=300.25\text{K}\)

Step2: Apply Boyle's Law (\(P_1V_1 = P_2V_2\)) for part (A)

Given \(P_1 = 0.929\text{atm}\), \(V_1=807\text{mL}\), \(P_2 = 0.634\text{atm}\)
We can solve for \(V_2\) using the formula \(V_2=\frac{P_1V_1}{P_2}\)
\(V_2=\frac{0.929\text{atm}\times807\text{mL}}{0.634\text{atm}}\)
\(V_2=\frac{750.703}{0.634}\text{mL}\approx1184\text{mL}\)

Step3: Apply Charles's Law (\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)) for part (B)

Given \(V_1 = 16.3\text{L}\), \(T_1=300.25\text{K}\), \(V_2 = 27.3\text{L}\)
We can solve for \(P_2\) using the formula \(P_2=\frac{V_1T_2}{V_2T_1}\times P_1\) (Since \(nR\) is constant in ideal gas law \(PV = nRT\), for same amount of gas \(n\) and \(R\) constant, \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), assume \(T_2=T_1\) (constant temperature) so \(P_2=\frac{V_1}{V_2}P_1\)
\(P_2=\frac{16.3\text{L}}{27.3\text{L}}\times0.859\text{atm}\)
\(P_2=\frac{14.0917}{27.3}\text{atm}\approx0.516\text{atm}\)

Answer:

(A) \(1184\) mL, (B) \(0.516\) atm