A gas occupies 8 litres at 23'C and 70cmHg what is the volume at stp

T1 = 273 + 23 = 300

T2 = 273 C standard
1 atm = 760 mm Hg = 76 cm Hg standard

n constant
P2 V2 / T2 = P1 V1 / T1
V2 = (P1/P2)(T2/T1) V1
V2 = (70/76)(273/300)8

PV = nRT

For the same amount of gas, n is constant. R is also constant.
So,

PV/T is constant.

=> (PV/T)1 = (PV/T)2

Plug in the given conditions: "8 litres at 23'C and 70cmHg" into the left side, and the STP temp and pressure into the right, to obtain for the latter volume.

Remember to convert Celsius to Kelvin!

A thread of mercury of length 15cm is used to trap some air in a capillary tube with uniform cross sectional area and closed at one end.With the tube vertical and the open and uppermost,the length of the trapped air column is rock.calculate the length of the air column when the tube is held(!)horizontally (!!)vertically with the open end underneath.atmospheric pressure =76cm of mercury

Well, if gas could speak, it would probably say something like, "Hey, don't pressure me!" But let's do some calculations instead.

To find the volume of the gas at STP (Standard Temperature and Pressure), we need to know the values of STP. It is typically defined as 0°C (273.15K) and 1 atmosphere (or 760 mmHg). Since the given gas is at 23°C and 70 cmHg, we need to make some conversions.

First, let's convert 70 cmHg to atm. We divide 70 by 760 (1 atm = 760 mmHg), which gives us approximately 0.092.

Next, we need to adjust the temperature from 23°C to Kelvin. We add 273.15 to 23, resulting in 296.15K.

Now, we can apply the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At STP, the pressure is 1 atm, the temperature is 296.15K, and the volume is unknown. If we assume the number of moles remains constant, we can use the equation P₁V₁/T₁ = P₂V₂/T₂, where the subscript ₁ represents the initial conditions and ₂ represents the final conditions.

Using this equation, we can solve for V₂, the volume at STP. Plugging in the values, we get:

(0.092)(8)/(296.15) = (1)(V₂)/(273.15)

Solving for V₂, we find that the volume at STP is approximately 0.248 L.

So, the gas would say, "I'm downsizing to 0.248 liters at STP! It's a gas-minute change!"

To find the volume of a gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law. The ideal gas law formula is:

PV = nRT

Where:
P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature of the gas in Kelvin

Now let's calculate the volume of the gas at STP.

Step 1: Convert the given temperature to Kelvin.

To convert Celsius to Kelvin, add 273.15 to the temperature.
Given temperature: 23°C

T = 23°C + 273.15 = 296.15 K

Step 2: Convert the pressure to atm.

The given pressure is in centimeters of mercury (cmHg). To convert it to atm, we divide the pressure by 760.

Given Pressure: 70 cmHg

P = 70 cmHg / 760 = 0.0921 atm (rounded to 4 decimal places)

Step 3: Plug the values into the ideal gas law equation to find the number of moles (n):

PV = nRT

V = 8 L (given)
P = 0.0921 atm
T = 296.15 K (converted)

n = (P * V) / (R * T)

n = (0.0921 atm * 8 L) / (0.0821 L·atm/(mol·K) * 296.15 K)

Step 4: Solve for the number of moles (n).

n ≈ 0.031 mol (rounded to 3 decimal places)

Step 5: Calculate the volume at STP.

Since we know the number of moles (n), we can use the equation:

V1 / T1 = V2 / T2

V1 = 8 L (given)
T1 = 296.15 K (converted)
V2 (volume at STP) = ?
T2 (STP temperature) = 273.15 K

Using the calculated value of n (0.031 mol) in the equation:

(8 L) / (296.15 K) = V2 / (273.15 K)

Cross-multiplying:

(8 L * 273.15 K) = (296.15 K * V2)

V2 = (8 L * 273.15 K) / (296.15 K)

Step 6: Solve for V2 (volume at STP).

V2 ≈ 7.37 L (rounded to 2 decimal places)

Therefore, the volume of the gas at STP is approximately 7.37 litres.