Question

question 4
v1 = 30 cu/ft
p1 =????? psi
t1 = 40°f

v2 = 45 cu/ft
p2 = 90 psi
t2 = 92°f

Explanation:

Step1: Recall the Ideal Gas Law for process

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), where temperatures must be in Rankine (R). First, convert Fahrenheit to Rankine: \(T = F + 459.67\). So \(T_1=40 + 459.67 = 499.67\,\text{R}\), \(T_2=92 + 459.67 = 551.67\,\text{R}\), \(V_1 = 30\,\text{cu/ft}\), \(V_2 = 45\,\text{cu/ft}\), \(P_2 = 90\,\text{psi}\).

Step2: Rearrange the formula to solve for \(P_1\)

From \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), we get \(P_1=\frac{P_2V_2T_1}{V_1T_2}\).

Step3: Substitute the values

Substitute \(P_2 = 90\), \(V_2 = 45\), \(T_1 = 499.67\), \(V_1 = 30\), \(T_2 = 551.67\) into the formula:
\(P_1=\frac{90\times45\times499.67}{30\times551.67}\)
First, calculate numerator: \(90\times45\times499.67 = 90\times22485.15 = 2023663.5\)
Denominator: \(30\times551.67 = 16550.1\)
Then \(P_1=\frac{2023663.5}{16550.1}\approx122.27\,\text{psi}\) (Wait, maybe miscalculation, let's recalculate more accurately. Wait, maybe I made a mistake in Rankine? Wait, actually, for ideal gas law, sometimes absolute temperature in Rankine or Kelvin. Let's check again. Wait, maybe the problem assumes using Fahrenheit directly? No, no, gas law needs absolute temperature. Wait, maybe the problem is using the combined gas law with \(T\) in Rankine. Wait, let's do exact calculation:

\(T_1 = 40 + 459.67 = 499.67\), \(T_2 = 92 + 459.67 = 551.67\)

\(P_1=\frac{90\times45\times499.67}{30\times551.67}=\frac{90\times45}{30}\times\frac{499.67}{551.67}= 135\times\frac{499.67}{551.67}\)

\(\frac{499.67}{551.67}\approx0.9057\), so \(135\times0.9057\approx122.27\). Wait, but maybe the problem expects using \(T\) in Fahrenheit? No, that's incorrect. But maybe a typo, or maybe the problem is using the formula with \(T\) in Fahrenheit (even though it's wrong, but let's check). If we use \(T\) in Fahrenheit:

\(P_1=\frac{90\times45\times40}{30\times92}=\frac{90\times45\times40}{2760}=\frac{162000}{2760}\approx58.69\). Wait, that's different. But the correct way is to use absolute temperature. Wait, maybe the problem is from a context where they use \(T\) in Rankine but approximate. Wait, maybe I messed up. Wait, let's check the formula again. The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), \(T\) in absolute scale. So Rankine is correct. So maybe my calculation is wrong. Wait, let's do it step by step:

\(90\times45 = 4050\); \(4050\times499.67 = 4050\times500 - 4050\times0.33 = 2025000 - 1336.5 = 2023663.5\)

\(30\times551.67 = 16550.1\)

\(2023663.5\div16550.1\approx122.27\). But maybe the problem has a different approach. Wait, maybe the user made a typo, but assuming the correct formula, let's proceed.

Wait, maybe I inverted the formula. Wait, no, the formula is correct. Wait, let's check with another approach. Let's use the formula \(\frac{P_1}{P_2}=\frac{V_2T_1}{V_1T_2}\) (from \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\) => \(\frac{P_1}{P_2}=\frac{V_2T_1}{V_1T_2}\)). So \(P_1 = P_2\times\frac{V_2T_1}{V_1T_2}\). So \(P_2 = 90\), \(V_2 = 45\), \(V_1 = 30\), \(T_1 = 499.67\), \(T_2 = 551.67\). So \(\frac{V_2}{V_1}=\frac{45}{30}=1.5\), \(\frac{T_1}{T_2}=\frac{499.67}{551.67}\approx0.9057\). Then \(P_1 = 90\times1.5\times0.9057\approx90\times1.35855\approx122.27\). So approximately 122 psi. But maybe the problem expects using \(T\) in Fahrenheit, which is wrong, but let's see: if \(T_1 = 40\), \(T_2 = 92\), then \(\frac{T_1}{T_2}=\frac{40}{92}\approx0.4348\), \(\frac{V_2}{V_1}=1.5\), so \(P_1 = 90\times1.5\times0.4348\approx90\times0.6522\approx58.7\). But that's incorrect because gas law needs absolute temperature. So the correct answer should be around 122 psi. Wait, maybe I made a mistake in Rankine? Wait, Rankine is Fahrenheit + 459.67, so 40 + 459.67 = 499.67, 92 + 459.67 = 551.67, correct. So the calculation is correct.

Answer:

\(P_1\approx122.27\,\text{psi}\) (or approximately 122 psi)