Question

at what temperature, in °c, would 1.13 mol of ideal gas occupy 54.7 l, at a pressure of 1.93 atm? report your answer to the nearest whole number.
r = 0.0821 \frac{l\cdot atm}{mol\cdot k} or 8.314 \frac{kpa\cdot l}{mol\cdot k}

Explanation:

Step1: Recall Ideal Gas Law

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant, and \( T \) is temperature (in Kelvin). We need to solve for \( T \), so rearrange the formula: \( T=\frac{PV}{nR} \).

Step2: Identify Values

Given: \( P = 1.93 \, \text{atm} \), \( V = 54.7 \, \text{L} \), \( n = 1.13 \, \text{mol} \), \( R = 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \) (since pressure is in atm, use this \( R \)).

Step3: Substitute Values into Formula

Plug in the values: \( T=\frac{(1.93 \, \text{atm})(54.7 \, \text{L})}{(1.13 \, \text{mol})(0.0821 \, \frac{\text{L·atm}}{\text{mol·K}})} \).

First, calculate numerator: \( 1.93 \times 54.7 \approx 105.571 \).

Then, calculate denominator: \( 1.13 \times 0.0821 \approx 0.092773 \).

Now, divide numerator by denominator: \( T \approx \frac{105.571}{0.092773} \approx 1138 \, \text{K} \) (in Kelvin).

Step4: Convert Kelvin to Celsius

Use the conversion \( ^\circ\text{C}=K - 273.15 \). So, \( ^\circ\text{C}=1138 - 273.15 \approx 865^\circ\text{C} \) (rounded to nearest whole number). Wait, wait, let's recalculate the Kelvin temperature more accurately.

Wait, let's redo the calculation of \( T \):

\( PV = 1.93 \times 54.7 = 1.93\times54.7 \). Let's compute 1.9354.7: 254.7=109.4, minus 0.07*54.7=3.829, so 109.4 - 3.829 = 105.571 (correct).

\( nR = 1.13\times0.0821 = 0.092773 \) (correct).

\( T = 105.571 / 0.092773 \approx 1138 \, \text{K} \) (exact division: 105.571 ÷ 0.092773 ≈ 1138). Then Celsius: 1138 - 273.15 = 864.85 ≈ 865°C. Wait, but maybe I made a mistake in calculation? Wait, let's check again.

Wait, 1.93 atm, 54.7 L, 1.13 mol, R=0.0821.

\( T = (1.93 54.7) / (1.13 0.0821) \)

Calculate 1.9354.7: 54.72 = 109.4, 54.7*0.07=3.829, so 109.4 - 3.829 = 105.571 (correct).

1.130.0821: 10.0821=0.0821, 0.13*0.0821=0.010673, total 0.0821+0.010673=0.092773 (correct).

105.571 / 0.092773 ≈ 1138 K (since 0.0927731138 ≈ 0.0927731000=92.773, 0.092773*138≈12.80, total 92.773+12.80≈105.573, which matches the numerator. So T≈1138 K.

Then Celsius: 1138 - 273 = 865 (since 273.15 is close to 273 for whole number, 1138-273=865). Wait, but maybe I messed up the gas constant? Wait, no, pressure is in atm, volume in L, moles in mol, so R=0.0821 is correct.

Wait, but let's check with more precise calculation:

1.93 54.7 = 1.93 54 + 1.93 * 0.7 = 104.22 + 1.351 = 105.571 (correct).

1.13 * 0.0821 = 0.092773 (correct).

105.571 / 0.092773 = let's do this division: 0.092773 * 1100 = 102.0503, 105.571 - 102.0503 = 3.5207. 3.5207 / 0.092773 ≈ 37.95. So total T≈1100 + 37.95≈1137.95 K≈1138 K. Then 1138 - 273.15 = 864.85≈865°C.

Answer:

\( 865^\circ\text{C} \) (Note: Wait, maybe I made a mistake in calculation? Wait, let's check again. Wait, 1.93 atm, 54.7 L, 1.13 mol. Let's use R=0.0821.

Wait, \( T = (PV)/(nR) = (1.93 54.7)/(1.13 0.0821) \)

1.93*54.7 = 105.571

1.13*0.0821 = 0.092773

105.571 / 0.092773 = 1138 K (approx). Then 1138 - 273 = 865. So the answer is 865.