Question

question 7
a 15.0 l container is filled with methane, ch₄, gas at 50.0 °c and 2.5 atm. what is the mass of the
methane in the container?
22.6 g
× incorrect. answer the question below and try again.
which gas law equation should be used to solve this problem?
p₁v₁ = p₂v₂
p₁/t₁ = p₂/t₂
v₁/t₁ = v₂/t₂
v₁/n₁ = v₂/n₂
pv = nrt
p₁v₁/t₁ = p₂v₂/t₂

Explanation:

Step1: Analyze the given information

We are given pressure \(P = 2.5\space atm\), volume \(V=15.0\space L\), temperature \(T = 50.0^{\circ}C=(50.0 + 273.15)\space K=323.15\space K\). We need to find the mass of \(CH_4\). The ideal gas law \(PV=nRT\) (where \(R = 0.0821\space L\cdot atm/(mol\cdot K)\)) relates \(P\), \(V\), \(n\) (moles), and \(T\). First, we solve for \(n\) (moles of \(CH_4\)) using \(PV=nRT\).

Step2: Rearrange the ideal gas law for \(n\)

From \(PV=nRT\), we can express \(n=\frac{PV}{RT}\). Substitute \(P = 2.5\space atm\), \(V = 15.0\space L\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), and \(T = 323.15\space K\) into the formula:

$$n=\frac{2.5\times15.0}{0.0821\times323.15}$$

$$n=\frac{37.5}{26.53}$$

$$n\approx1.41\space mol$$

Step3: Calculate the mass of \(CH_4\)

The molar mass of \(CH_4\) is \(M=(12.01+4\times1.008)\space g/mol = 16.04\space g/mol\). Using the formula \(m=n\times M\), substitute \(n = 1.41\space mol\) and \(M = 16.04\space g/mol\)

$$m=1.41\times16.04$$

$$m\approx22.6\space g$$

Answer:

\(PV = nRT\)