If four moles of a gas of 5.4 atmospheres have a volume of 120 liters, what is the temperature?

Question

Peyton Adams · Accepted Answer

#T=(PV)/(nR)#, i.e. the Ideal Gas equation, and we have all the necessary parameters. We get an answer in degrees Kelvin not Centigrade.

Chemists generally use the gas constant #R#, where #R=0.0821*L*atm*K^-1mol^-1# (why? because chemists typically use litre volumes, and 1 atmosphere is easy to measure by means of a mercury column). #T# #=# #(5.4*cancel(atm)xx120*cancelL)/(4*cancel(mol)xx0.0821*cancelL*cancel(atm)*K^-1cancel(mol^-1))# #=# #(????)/(K^-1)# #=# #??K# So the answer, whatever it is, is in degrees #K", absolute temperature"# as required. Remember that #"degrees Kelvin, K"# #=# #""^@C+273.15#.

Natalie Anton · Answer

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

$T = \frac{{PV}}{{nR}}$

Given: 
$P = 5.4 \text{ atm}$
$V = 120 \text{ L}$
$n = 4 \text{ moles}$
$R = 0.0821 \text{ atm} \cdot \text{L/mol} \cdot \text{K}$

Substituting the values into the equation:

$T = \frac{{5.4 \times 120}}{{4 \times 0.0821}}$

$T ≈ 195.32 \text{ K}$