Question

question 5
v1 = 300 cu/ft
p1 = 14.7 psi
t1 = 59°f
v2 = 300 cu/ft
p2 = 4.35 psi
t2 =???????°f

Explanation:

Step1: Identify the Gas Law

Since \( V_1 = V_2 = 300\ \text{cu/ft} \) (volume is constant), we use Gay - Lussac's Law (or Pressure - Temperature Law) for a fixed mass of gas at constant volume: \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), where temperatures must be in Rankine (\( T(\text{R})=T(\text{F}) + 459.67\)).

First, convert \( T_1 \) to Rankine:
\( T_{1,\text{R}}=59 + 459.67=518.67\ \text{R} \)

Step2: Rearrange the Formula for \( T_2 \)

From \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), we solve for \( T_2 \):
\( T_2=\frac{P_2\times T_1}{P_1} \)

Substitute \( P_1 = 14.7\ \text{psi} \), \( P_2 = 4.35\ \text{psi} \), and \( T_{1,\text{R}} = 518.67\ \text{R} \):
\( T_{2,\text{R}}=\frac{4.35\times518.67}{14.7} \)

Calculate the numerator: \( 4.35\times518.67\approx4.35\times518.67 = 2256.2145 \)

Then divide by \( 14.7 \): \( T_{2,\text{R}}\approx\frac{2256.2145}{14.7}\approx153.48\ \text{R} \)

Step3: Convert \( T_2 \) back to Fahrenheit

Use \( T(\text{F})=T(\text{R}) - 459.67 \):
\( T_2=153.48 - 459.67\approx - 306.19^\circ\text{F} \) (Wait, this seems odd. Wait, maybe we made a mistake in the law? Wait, if volume is constant (\( V_1 = V_2 \)), then \( \frac{P_1}{T_1}=\frac{P_2}{T_2} \) holds. But let's check the calculation again.

Wait, recalculate \( T_{2,\text{R}} \):
\( \frac{4.35\times518.67}{14.7}=\frac{4.35}{14.7}\times518.67\approx0.2959\times518.67\approx153.48\ \text{R} \). Then \( T_2(\text{F})=153.48 - 459.67\approx - 306.19^\circ\text{F} \). But let's verify the law again. Gay - Lussac's Law: \( \frac{P_1}{T_1}=\frac{P_2}{T_2} \) (constant \( V \), \( n \)). So the formula is correct. Maybe the problem has a typo, but following the math:

Wait, maybe we should use \( T(\text{R})=T(\text{F})+459.67 \), so \( T_1 = 59^\circ\text{F}=59 + 459.67 = 518.67\ \text{R} \), \( P_1 = 14.7\ \text{psi} \), \( P_2 = 4.35\ \text{psi} \). Then \( T_2=\frac{P_2\times T_1}{P_1}=\frac{4.35\times518.67}{14.7}\approx153.48\ \text{R} \), then \( T_2(\text{F})=153.48 - 459.67\approx - 306.2^\circ\text{F} \)

Answer:

\( T_2\approx\boldsymbol{-306.2^\circ\text{F}} \) (Note: The negative temperature suggests a very low temperature, which might be correct for a gas under these pressure changes at constant volume, but it's unusual. Check the problem's given values again for accuracy.)