A sample of an ideal gas has a volume of 2.34 L at 279 K and 1.10 atm. What is the pressure when the volume is 1.78 L and the temperature is 304 K?

Answer 1

Pressure = #1.6 atm#

Use the formula #(P_1V_1)/T_1 = (P_2V_2)/T_2#, where #P_1# is the initial pressure, #V_1# is the initial volume, and #T_1# is the initial temperature. And #V_2\, P_2\, T_2# are the final values (with the variable you're trying to find!)

Enter the values as replacements in the formula.

#((1.10atm)(2.34L))/(279K)= ((P_2)(1.78L))/(304K)#

#(304)(1.10)(2.34) = (P_2)(1.78)(279)#

#782.5 = 496.6 (P_2)#

#P_2 = 1.6 atm#

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Answer 2

Cooper Adams

Using the ideal gas law equation, $PV = nRT$ , where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is temperature in Kelvin, we can solve for the new pressure:

$P_1V_1 / T_1 = P_2V_2 / T_2$

$P_2 = (P_1V_1T_2) / (V_2T_1)$

Given:
$P_1 = 1.10 \, \text{atm}$ ,
$V_1 = 2.34 \, \text{L}$ ,
$T_1 = 279 \, \text{K}$ ,
$V_2 = 1.78 \, \text{L}$ ,
$T_2 = 304 \, \text{K}$

$P_2 = (1.10 \times 2.34 \times 304) / (1.78 \times 279)$

$P_2 ≈ 1.76 \, \text{atm}$

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.