Question

A sample of an ideal gas has a volume of 2.26 L at 289 K and 1.03 atm. How do you calculate the pressure when the volume is 1.64 L and the temperature is 308 K?

Answer 1

The ideal gas law is the equation of state for an ideal gas (hypothetical). It is a good approximation of the behavior of many gases under certain conditions. It is a combination of the empirical Charles's law, Boyle's law, Avogadro's law, and Gay-Lussac's law.
Formula used:
For constant no. of moles of ideal gas: $\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$

Complete answer:
The empirical relationship between volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law which can be written as follows:-
$PV=nRT$
Where,
P = pressure of the gas
V =volume of gas it occupies
n = number of moles of gas present in the solution or sample
R = universal gas constant, equal to $0.0821\dfrac{atm\cdot L}{mol\cdot K}$
T = absolute temperature of the gas
When the quantity of gas is constant i.e., no. of moles is same then:-
${{n}_{1}}={{n}_{2}}$
$\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$
The values are as follows:
$\begin{aligned} & {{P}_{1}}=1.03atm\text{ }{{P}_{2}}=? \\\ & {{V}_{1}}=2.26L\text{ }{{\text{V}}_{2}}=1.64L \\\ & {{T}_{1}}=289K\text{ }{{\text{T}}_{2}}=308K \\\ \end{aligned}$
On substituting all these values in the formula, we get:
$\dfrac{1.03atm\times 2.26L}{289K}=\dfrac{{{P}_{2}}\times 1.64L}{308K}$
$\dfrac{1.03atm\times 2.26L\times 308K}{289K\times 1.64L}={{P}_{2}}$
${{P}_{2}}=1.51atm$
Therefore, the pressure is 1.51 atm when the volume is 1.64 L and the temperature is 308 K for the sample of an ideal gas.

Note:
Before solving any question, carefully write down the values provided in the question. Always check the units and in order to avoid mistakes, try to solve the entire question along with units (preferably convert all the values to the same units before moving to further calculations).