Question

v1 = 32 cu/ft
p1 =?????? psi
t1 = 50°f

v2 = 40 cu/ft
p2 = 90 psi
t2 = 85°f

Explanation:

Step1: Recall the Ideal Gas Law for Initial and Final States

The ideal gas law is \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \) (where temperatures are in absolute units, but since we are taking the ratio, we can use Rankine or convert Fahrenheit to Rankine by \( R = F + 459.67 \). However, when taking the ratio \( \frac{T_1}{T_2} \) (in Rankine) will have the same ratio as \( \frac{T_1 + 459.67}{T_2 + 459.67} \) but actually, for the combined gas law, we should use absolute temperature. Let's convert \( T_1 = 50^\circ F \) to Rankine: \( T_{R1}=50 + 459.67 = 509.67\ R \), \( T_{R2}=85 + 459.67 = 644.67\ R \). But we can also use the formula \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \) with \( T \) in Rankine (or we can keep it as Fahrenheit in the ratio? Wait, no, the temperature must be in absolute scale (Rankine for Fahrenheit, Kelvin for Celsius). So first, convert temperatures to Rankine.

Step2: Rearrange the Formula to Solve for \( P_1 \)

From \( \frac{P_1V_1}{T_{R1}}=\frac{P_2V_2}{T_{R2}} \), we can solve for \( P_1 \):
\( P_1=\frac{P_2V_2T_{R1}}{V_1T_{R2}} \)

Step3: Substitute the Values

We have \( P_2 = 90\ psi \), \( V_2 = 40\ cu/ft \), \( T_{R1}=50 + 459.67 = 509.67\ R \), \( V_1 = 32\ cu/ft \), \( T_{R2}=85 + 459.67 = 644.67\ R \)

Substitute into the formula:
\( P_1=\frac{90\times40\times509.67}{32\times644.67} \)

First, calculate numerator: \( 90\times40 = 3600 \); \( 3600\times509.67 = 3600\times500 + 3600\times9.67 = 1,800,000 + 34,812 = 1,834,812 \)

Denominator: \( 32\times644.67 = 20,629.44 \)

Then \( P_1=\frac{1,834,812}{20,629.44}\approx 88.94\ psi \) (Wait, but let's check the calculation again. Wait, maybe I made a mistake in the temperature conversion. Wait, actually, the combined gas law is \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \) where \( T \) is in absolute temperature (Rankine or Kelvin). So let's do the calculation with more precise steps.

Alternatively, maybe the problem assumes that the temperature is in Rankine, but let's use the formula:

\( P_1=\frac{P_2V_2T_1}{V_1T_2} \) (Wait, no, that's wrong. The correct formula is \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \) => \( P_1 = P_2\times\frac{V_2}{V_1}\times\frac{T_1}{T_2} \), but \( T_1 \) and \( T_2 \) must be in absolute units. So \( T_1 = 50 + 459.67 = 509.67\ R \), \( T_2 = 85 + 459.67 = 644.67\ R \)

So \( \frac{T_1}{T_2}=\frac{509.67}{644.67}\approx0.7906 \)

\( \frac{V_2}{V_1}=\frac{40}{32}=1.25 \)

\( P_1 = 90\times1.25\times0.7906 \)

Calculate \( 90\times1.25 = 112.5 \)

Then \( 112.5\times0.7906\approx 88.94\ psi \)

Wait, but maybe the problem is using the combined gas law with temperature in Fahrenheit? No, that's incorrect. But maybe it's a simplified problem where they consider the ratio of temperatures as \( \frac{T_1 + 460}{T_2 + 460} \) (using 460 instead of 459.67 for approximation). Let's try that:

\( T_1 = 50 + 460 = 510\ R \), \( T_2 = 85 + 460 = 645\ R \)

\( \frac{T_1}{T_2}=\frac{510}{645}=\frac{34}{43}\approx0.7907 \)

\( \frac{V_2}{V_1}=\frac{40}{32}=\frac{5}{4}=1.25 \)

\( P_1 = 90\times\frac{5}{4}\times\frac{34}{43} \)

Calculate \( 90\times\frac{5}{4}=112.5 \)

\( 112.5\times\frac{34}{43}=\frac{112.5\times34}{43}=\frac{3825}{43}\approx88.95\ psi \)

So approximately 89 psi. But let's check the formula again. The combined gas law is \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \), so solving for \( P_1 \):

\( P_1 = \frac{P_2V_2T_1}{V_1T_2} \) (with \( T \) in absolute units)

Yes, that's the correct rearrangement. So substituting the values:

\( P_1 = \frac{90\times40\times509.67}{32\times644.67} \)

Calculate numerator: \( 90\times40 = 3600 \); \( 3600\times509.67 = 3600\times500 + 3600\times9.67 = 1,800,000 + 34,812 = 1,834,812 \)

Denominator: \( 32\times644.67 = 20,629.44 \)

\( P_1 = \frac{1,834,812}{20,629.44}\approx88.94\ psi \approx 89\ psi \)

Answer:

\( P_1\approx\boxed{89}\ psi \) (or more precisely, approximately 88.9 psi, but rounded to a reasonable decimal place)