A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas?

Question

Serenity Allen · Accepted Answer

The sample of gas contains 1.3 moles of gas.

Use the ideal gas law as shown below.

#P="121.59 kPa"#
#V="31 L"#
#T="360 K"#
#R=8.3144598 "L kPa K"^(-1) "mol"^(-1)"#

Rearrange the equation to isolate #n#, plug in the known values, and solve.

#n=(PV)/(RT)#

#n=(121.59cancel"kPa"xx31cancel"L")/(8.3144598cancel"L kPa K"^(-1) "mol"^(-1)xx360cancel"K")="1.3 mol"# (rounded to two significant figures)

Xavier Avery · Answer

undefined You can use the ideal gas law, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

Given:
Pressure ($P$) = 121.59 kPa
Volume ($V$) = 31 L
Temperature ($T$) = 360 K

Rearrange the ideal gas law to solve for $n$:
$$ n = \frac{PV}{RT} $$

Substitute the given values into the equation:
$$ n = \frac{(121.59 \, \text{kPa}) \times (31 \, \text{L})}{(8.314 \, \text{L} \cdot \text{kPa/mol} \cdot \text{K}) \times (360 \, \text{K})} $$

$$ n \approx \frac{(3772.29 \, \text{L} \cdot \text{kPa})}{(2990.64 \, \text{L} \cdot \text{kPa/mol})} $$

$$ n \approx 1.26 \, \text{mol} $$

Therefore, the sample of gas contains approximately 1.26 moles of gas.