Question

question 2
4 pts
v1 = 105 cu/ft
p1 = 180 psi
t1 = 60°f

v2 = 200 cu/ft
p2 = 320 psi
t2 =???????°f

Explanation:

Step1: Recall Ideal Gas Law for Process

For a gas undergoing a process, we can use the combined gas law (derived from ideal gas law \( PV = nRT \), assuming \( n \) is constant): \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \). We need to solve for \( T_2 \). First, rearrange the formula to \( T_2=\frac{P_2V_2T_1}{P_1V_1} \).

Step2: Convert Temperature to Rankine (since gas law uses absolute temp)

\( T_1 = 60^\circ\text{F}+ 459.67 = 519.67\,\text{R} \) (we can use 460 for approximation: \( 60 + 460 = 520\,\text{R} \) for simplicity).

Step3: Substitute Values

\( P_1 = 180\,\text{psi} \), \( V_1 = 105\,\text{cu/ft} \), \( P_2 = 320\,\text{psi} \), \( V_2 = 200\,\text{cu/ft} \), \( T_1 = 520\,\text{R} \).

\( T_2=\frac{320\times200\times520}{180\times105} \)

First, calculate numerator: \( 320\times200 = 64000 \); \( 64000\times520 = 33280000 \)

Denominator: \( 180\times105 = 18900 \)

Then, \( T_2=\frac{33280000}{18900}\approx1760.85\,\text{R} \)

Step4: Convert Back to Fahrenheit

\( T_2 (\text{F})= T_2 (\text{R}) - 459.67 \approx1760.85 - 459.67\approx1301.18^\circ\text{F} \) (using 460: \( 1760.85 - 460 = 1300.85^\circ\text{F} \), close enough)

Wait, wait, maybe I made a mistake in assuming ideal gas? Wait, the problem might be using the combined gas law correctly. Wait, let's recalculate with exact values:

\( T_1 = 60 + 459.67 = 519.67\,\text{R} \)

\( T_2=\frac{320\times200\times519.67}{180\times105} \)

Numerator: \( 320\times200 = 64000 \); \( 64000\times519.67 = 64000\times500 + 64000\times19.67 = 32000000 + 1258880 = 33258880 \)

Denominator: \( 180\times105 = 18900 \)

\( T_2=\frac{33258880}{18900}\approx1759.73\,\text{R} \)

\( T_2 (\text{F})= 1759.73 - 459.67\approx1300.06^\circ\text{F} \approx1300^\circ\text{F} \) (or more precise)

Wait, maybe the problem expects using the combined gas law directly without Rankine? No, gas laws require absolute temperature (Rankine for Fahrenheit, Kelvin for Celsius). So the key steps are using \( \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \), converting T1 to Rankine, solving for T2 (Rankine), then converting back to Fahrenheit.

Answer:

\( \approx1300^\circ\text{F} \) (or more precisely ~1301°F)