# Ideal Gas Law and Pressure at 80°C

• SakuRERE
In summary, the correct way to solve this problem is to first find the volume of the gas using the given molar mass and density, and then substitute that value into the ideal gas law. Your professor did this correctly, while your answer is inconsistent due to rounding errors. It is important to be consistent with significant figures when rounding final answers.
SakuRERE

## Homework Statement

An ideal gas has a molar mass of 40 g and a density of 1.2 kg m-3 at 80°C. What is its pressure at that temperature?

## Homework Equations

PV=nRT
R constant= 8.314
n= number of moles
T= tempreture in kelvin
density=Mass/ Volume

## The Attempt at a Solution

i simply solved it like this:
i found first the volume:
V=molar mass/ density
V= 0.04kg/ 1.2kg/m^3
= 0.03 m^3
and then i just substituted in the ideal gas law

P=nRT/V (where n is the moles and here we have 1 mole)
P= (1)*(8.314)*(353K)/ (0.03) = 97828.06 Pa

but my professor solved it like this:
T = 273 + 80 = 353 K

PV = nRT

m = rV

so V = 0.04 / 1.2 with n = 1 (since 1 mole of gas has a mass of 40 g, or 0.04 kg)

P = RT/V = 8.3 x 353 x 1.2/0.04 = 8.79x104 Pa

I noticed that he didn't find the volume first and then substituted but he substituted directly in the ideal gas law! why it should be like this!?

SakuRERE said:

## Homework Statement

An ideal gas has a molar mass of 40 g and a density of 1.2 kg m-3 at 80°C. What is its pressure at that temperature?

## Homework Equations

PV=nRT
R constant= 8.314
n= number of moles
T= tempreture in kelvin
density=Mass/ Volume

## The Attempt at a Solution

i simply solved it like this:
i found first the volume:
V=molar mass/ density
V= 0.04kg/ 1.2kg/m^3
= 0.03 m^3
and then i just substituted in the ideal gas law

P=nRT/V (where n is the moles and here we have 1 mole)
P= (1)*(8.314)*(353K)/ (0.03) = 97828.06 Pa

but my professor solved it like this:
T = 273 + 80 = 353 K

PV = nRT

m = rV

so V = 0.04 / 1.2 with n = 1 (since 1 mole of gas has a mass of 40 g, or 0.04 kg)

P = RT/V = 8.3 x 353 x 1.2/0.04 = 8.79x104 Pa

I noticed that he didn't find the volume first and then substituted but he substituted directly in the ideal gas law! why it should be like this!?
Because your professor did it right. He substituted values after he derived the algebraic expression for the unknown quantity in terms of the given quantities. Your answer is inconsistent with your professor's (and the correct) answer because you rounded the volume from 0.033333 m3 to 0.03 m3.

SakuRERE
And you rounded the volume to 1 significant figure, but quoted the answer to 7 sig figs! The professor gave his answer to 3 sig figs, consistent with the data provided.
Rule: do the calculations to high accuracy, round the final answer to the appropriate number of sig figs.

SakuRERE

## 1. What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

## 2. How does the ideal gas law relate to pressure?

The ideal gas law states that pressure and volume are inversely proportional, meaning that as one increases, the other decreases, assuming all other variables remain constant. This is known as Boyle's Law.

## 3. What factors affect pressure according to the ideal gas law?

The ideal gas law shows that pressure is affected by the volume, temperature, and number of moles of an ideal gas. As volume decreases, pressure increases, and as temperature increases, pressure also increases. Additionally, an increase in the number of moles will result in an increase in pressure.

## 4. How is pressure measured in the ideal gas law?

In the ideal gas law, pressure is typically measured in units of atmospheres (atm), but it can also be measured in other units such as Pascals (Pa) or kilopascals (kPa). It is important to use the appropriate units for each variable in the equation to ensure accurate calculations.

## 5. Can the ideal gas law be applied to real gases?

The ideal gas law is an approximation and is most accurate for ideal gases, which do not exist in the real world. However, it can still be used to make predictions and calculations for real gases under certain conditions, such as low pressures and high temperatures.

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