Question

question
a 50.0 l container holds 45.0 moles of gas. the temperature of the container is 200.°c. what is the pressure?

• use ( r = 0.08206 \frac{l atm}{mol k} ) for the gas constant.
• use ( -273.15^{circ}c ) for absolute zero.

select the correct answer below:
23.7 atm
34.9 atm
41.8 atm
50.6 atm

Explanation:

Step1: Convert temperature to Kelvin

$$T=200.^{\circ}C + 273.15=473.15\,\text{K}$$

Step2: Use ideal gas law \(PV = nRT\) to solve for \(P\)

We know \(V = 50.0\,\text{L}\), \(n=45.0\,\text{mol}\), \(R = 0.08206\frac{\text{L atm}}{\text{mol K}}\), \(T = 473.15\,\text{K}\)
From \(PV=nRT\), we can express \(P=\frac{nRT}{V}\)
Substitute the values:
$$P=\frac{45.0\,\text{mol}\times0.08206\frac{\text{L atm}}{\text{mol K}}\times473.15\,\text{K}}{50.0\,\text{L}}$$
First calculate the numerator: \(45.0\times0.08206\times473.15 = 45.0\times38.817\approx1746.765\)
Then divide by \(V = 50.0\): \(P=\frac{1746.765}{50.0}=34.9353\approx34.9\,\text{atm}\)

Answer:

34.9 atm