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a gas has an initial volume of 24.6 l at a pressure of 1.90 atm and a t…

Question

a gas has an initial volume of 24.6 l at a pressure of 1.90 atm and a temperature of 335 k. the pressure of the gas increases to 3.50 atm, and the volume of the gas increases to 31.3 l. what is the final temperature of the gas? 143 k 231 k 485 k 785 k

Explanation:

Step1: Recall the Combined Gas Law

The combined gas law is given by \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), where \(P_1, V_1, T_1\) are initial pressure, volume, temperature and \(P_2, V_2, T_2\) are final pressure, volume, temperature.

Step2: Identify the given values

\(P_1 = 1.90\) atm, \(V_1=24.6\) L, \(T_1 = 335\) K, \(P_2=3.50\) atm, \(V_2 = 31.3\) L. We need to find \(T_2\).

Step3: Rearrange the formula to solve for \(T_2\)

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

Step4: Substitute the values into the formula

\(T_2=\frac{3.50\times31.3\times335}{1.90\times24.6}\)
First, calculate the numerator: \(3.50\times31.3 = 109.55\), then \(109.55\times335=109.55\times300 + 109.55\times35=32865+3834.25 = 36699.25\)
Then, calculate the denominator: \(1.90\times24.6 = 46.74\)
Now, divide numerator by denominator: \(T_2=\frac{36699.25}{46.74}\approx785\) K? Wait, no, wait, let's recalculate:
Wait, \(3.50\times31.3 = 109.55\), \(109.55\times335\): \(109.55\times300 = 32865\), \(109.55\times35=3834.25\), sum is \(32865 + 3834.25=36699.25\)
Denominator: \(1.90\times24.6 = 46.74\)
\(36699.25\div46.74\approx785\)? Wait, no, wait, maybe I made a miscalculation. Wait, let's check again:
Wait, \(3.50\times31.3 = 109.55\), \(109.55\times335\): Let's do \(335\times100 = 33500\), \(335\times9.55=335\times(9 + 0.55)=3015+184.25 = 3199.25\), so total numerator is \(33500+3199.25 = 36699.25\)
Denominator: \(1.90\times24.6 = 46.74\)
\(36699.25\div46.74\approx785\)? Wait, but let's check the calculation again. Wait, maybe I messed up the numbers. Wait, \(3.50\times31.3 = 109.55\), \(1.90\times24.6 = 46.74\), \(109.55\div46.74\approx2.344\), then \(2.344\times335\approx2.344\times300=703.2\), \(2.344\times35 = 82.04\), total \(703.2 + 82.04 = 785.24\approx785\) K? Wait, but let's check the options. Wait, maybe I made a mistake. Wait, no, let's recalculate:

Wait, \(P_2 = 3.50\), \(V_2 = 31.3\), \(T_1 = 335\), \(P_1 = 1.90\), \(V_1 = 24.6\)

\(T_2=\frac{3.50\times31.3\times335}{1.90\times24.6}\)

Calculate numerator: \(3.50\times31.3 = 109.55\); \(109.55\times335 = 109.55\times(300 + 35)=109.55\times300+109.55\times35 = 32865+3834.25 = 36699.25\)

Denominator: \(1.90\times24.6 = 46.74\)

\(T_2=\frac{36699.25}{46.74}\approx785\) K

Answer:

785 K (the option with 785 K)