Question

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if 1.50 mol has a pressure of 1.25 atm at a temperature of $-6^{\circ }\mathrm{C};\mathbf{b}$ ) the absolute temperature of the gas at which $3.33\times {10}^{-3}$ mol occupies 478 $\mathrm{mL}$ at 750 torr; (c) the pressure, in atmospheres, if 0.00245 $\mathrm{mol}$ occupies 413 $\mathrm{mL}$ at ${138}^{\circ }\mathrm{C};(\mathbf{d})$ the quantity of gas, in moles, if 126.5 $\mathrm{L}$ at ${54}^{\circ }\mathrm{C}$ has a pressure of 11.25 $\mathrm{kPa}$ .

Short answer

The short answer for each part is as follows: (a) The volume of the gas is $16.54L$. (b) The absolute temperature of the gas is $170.90K$. (c) The pressure of the gas is $0.487atm$. (d) The quantity of gas is $5.92mol$.

Step-by-step solution

Part (a): Calculate the volume of the gas

We are given the pressure $P=1.25atm$, the number of moles $n=1.50mol$, and the temperature $T=-6^{\circ }C$. We have to find the volume of the gas (V). First, we need to convert the temperature into Kelvin (K). Temperature in Kelvin (K) = Temperature in Celsius (C) + 273.15 $T=-6^{\circ }C+273.15=267.15K$ Now we can use the ideal gas law equation $PV=nRT$ to find the volume. Here, we will use the value of R in atm L/mol K which is 0.08206 $atmL/(molK)$. $V=\frac{nRT}{P}=\frac{1.50mol\times 0.08206\frac{atmL}{molK}\times 267.15K}{1.25atm}=16.54L$ The volume of the gas is 16.54 liters.

Part (b): Calculate the absolute temperature of the gas

We are given the number of moles $n=3.33\times {10}^{-3}mol$, the volume V = 478 mL, and the pressure $P=750torr$. We need to find the temperature of the gas (T). First, we need to convert the pressure to atm and the volume to L: $P=\frac{750torr}{760torr/atm}=0.9868atm$ $V=\frac{478mL}{1000mL/L}=0.478L$ Now we can use the ideal gas law equation $PV=nRT$ to find the temperature. $T=\frac{PV}{nR}=\frac{0.9868atm\times 0.478L}{3.33\times {10}^{-3}mol\times 0.08206\frac{atmL}{molK}}=170.90K$ The absolute temperature of the gas is 170.90 K.

Part (c): Calculate the pressure of the gas

We are given the number of moles $n=0.00245mol$, the volume $V=413mL$, and the temperature $T={138}^{\circ }C$. We have to find the pressure of the gas (P). First, convert the temperature to Kelvin and the volume to L: $T={138}^{\circ }C+273.15=411.15K$ $V=\frac{413mL}{1000mL/L}=0.413L$ Now we can use the ideal gas law equation $PV=nRT$ to find the pressure. $P=\frac{nRT}{V}=\frac{0.00245mol\times 0.08206\frac{atmL}{molK}\times 411.15K}{0.413L}=0.487atm$ The pressure of the gas is 0.487 atmospheres.

Part (d): Calculate the quantity of gas in moles

We are given the volume $V=126.5L$, the temperature $T={54}^{\circ }C$, and the pressure $P=11.25kPa$. We have to find the number of moles (n) of the gas. First, we need to convert the temperature to Kelvin and the pressure to atm: $T={54}^{\circ }C+273.15=327.15K$ $P=\frac{11.25kPa}{101.325kPa/atm}=0.1110atm$ Now we can use the ideal gas law equation $PV=nRT$ to find the number of moles. $n=\frac{PV}{RT}=\frac{0.1110atm\times 126.5L}{0.08206\frac{atmL}{molK}\times 327.15K}=5.92mol$ The quantity of gas is 5.92 moles.

Key concepts

PV=nRT

The ideal gas law is a fundamental equation in chemistry and physics that links together four key quantities of a gas: pressure (P), volume (V), number of moles (n), and temperature (T). The equation is expressed as $PV=nRT$. Here, R is the ideal gas constant, which has a value of 0.08206 atm L/mol K when using these units.

By rearranging the equation, you can solve for any one of these variables if the others are known. For example, to find volume, you can use $V=\frac{nRT}{P}$. This makes the ideal gas law a powerful tool for solving problems involving gases, as you can calculate pressure, volume, temperature, or the amount of substance in moles given enough information about the other variables.

• Understand that $n$ is the number of moles.
• R = 0.08206 atm L/mol K is the constant used in these calculations.
• Ensure all units are appropriate before inserting into the equation.

moles calculation

In chemistry, calculating moles is essential when dealing with gas problems. Moles represent the amount of substance present. We can calculate moles using the ideal gas law by rearranging it to $n=\frac{PV}{RT}$.

This is useful when you know the pressure, volume, and temperature of a gas. Just plug these values into the equation with the proper units, and you can find out how much gas is present in moles.

• Use the ideal gas law rearranged for moles: $n=\frac{PV}{RT}$.
• Ensure your data (P, V, T) is in the correct units before calculating.

Moles act as a bridge allowing us to connect chemical equations with physical quantities of gases.

temperature conversion

Temperature plays a critical role in gas calculations. In these equations, temperature must always be expressed in Kelvin (K). This is because Kelvin is an absolute temperature scale, and the relationships in the ideal gas law depend on absolute temperature.

Conversion from Celsius to Kelvin is straightforward. Simply add 273.15 to the Celsius temperature: $T_K=T_C+273.15$.

• Always convert °C to K before solving gas law problems.
• Remember: Kelvin = Celsius + 273.15.

When plugging values into the ideal gas law, using Kelvin ensures the accuracy of your calculations and results.

pressure conversion

Pressure must be converted to consistent units before performing calculations with the ideal gas law. Often, pressure needs to be in atmospheres (atm) for the constant R (0.08206 atm L/mol K) to be applicable.

For example, to convert from torr to atm, use the relationship $1\text{atm}=760\text{torr}$. To convert kPa to atm, recall $1\text{atm}=101.325\text{kPa}$.

• Use $P(\text{atm})=\frac{P(\text{torr})}{760}$ to convert torr to atm.
• Use $P(\text{atm})=\frac{P(\text{kPa})}{101.325}$ for kPa to atm conversion.

Converting pressures correctly ensures that the calculations using the ideal gas law remain accurate and reliable.