Question

if 3.62 mol of an ideal gas has a pressure of 2.33 atm and a volume of 76.35 l, what is the temperature of the sample in degrees celsius? t = °c

Explanation:

Step1: Use ideal gas law

The ideal gas law is \(PV = nRT\), where \(P = 2.33\ atm\), \(V=76.35\ L\), \(n = 3.62\ mol\), and \(R=0.0821\ L\cdot atm/(mol\cdot K)\). Rearrange for \(T\): \(T=\frac{PV}{nR}\).

Step2: Substitute values

Substitute the values: \(T=\frac{2.33\times76.35}{3.62\times0.0821}\).
First, calculate the numerator: \(2.33\times76.35 = 177.8955\).
Then, calculate the denominator: \(3.62\times0.0821=0.297202\).
So, \(T=\frac{177.8955}{0.297202}\approx598.6\ K\).

Step3: Convert to Celsius

Use the conversion \(T_{C}=T_{K}-273.15\).
\(T_{C}=598.6 - 273.15=325.45^{\circ}C\).

Answer:

\(325^{\circ}C\)