Combined Gas Law Calculator

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**$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$**

In the combined gas laws the ratio of $\frac{PV}{T}$ equals a constant for a final amount of gas. The value of this constant is actually proportional to the number of moles of gas n.

**Step 1 :**

**Step 2 :**

The relationship among pressure, temperature and volume of a gas sample can be expressed in a mathematical formula called the combined gas law. This law allows to predict how any of these properties of a gas will change if we change one or both of the other two.

For example, if we change the temperature of the gas inside a rigid steel cylinder, we can predict how the pressure of the gas will change.

For example, if we change the temperature of the gas inside a rigid steel cylinder, we can predict how the pressure of the gas will change.

The mathematical equation for the combined gas law is

In the combined gas laws the ratio of $\frac{PV}{T}$ equals a constant for a final amount of gas. The value of this constant is actually proportional to the number of moles of gas n.

Read te problem and list out the values given.

Substitute the given values in the corresponding formula to find out the unknown value.

**$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$**

In an autoclave, steam at 100

^{o}C generated at 1.00atm. After the autoclave is closed the steam is heated at constant volume until the pressure gauge indicates 1.13atm. What is the final temperature in the autoclave?**Step 1 :**Given data

P

_{1}= 1.00atm

P_{2}= 1.13atm

T_{1}= 100^{o}C = 100 + 273 = 373K

V

_{1}and V_{2}are the same in this example and consequentyly cancel each other.**Step 2 :**Substitute the values in the corresponding formula to obtain the T

_{2}value.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

$\frac{1.00 \times}{373} = \frac{1.13}{T_{2}}$

T

_{2}= 421K**Answer :**The final temperature is T

_{2}= 421Kor

T

_{2}= 421 - 273 = 148^{o}CA gas in a fexible container has a voume of 0.50L and a pressure of 1.0atm at 393K. When the gas id heated to 500K its volume expands to 3.0L. What is the new pressure of the gas in the flexibe container?

**Step 1 :**Given data

P

_{1}= 1.0atm

V_{1}= 0.50L

V_{2}= 3.0L

T_{1}= 393K

T_{2}= 500K

P

_{2}= ?**Step 2 :**Substitute the values in the corresponding formula to obtain the P

_{2}value.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

$\frac{1.0 \times 0.50}{393} = \frac{P_{2}\times3.0}{500}$

P

_{2}= 0.21atm**Answer :**The pressure P

_{2}= 0.21atm