3 Liter of a gas is collected at 35 degrees celcius and 705 mmHg .what is the volume at STP?

(P1V1/T1) = (P2V2/T2)

To find the volume of the gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation, which states:

PV = nRT

where:
P = pressure of the gas (given in mmHg)
V = volume of the gas (unknown)
n = number of moles of gas
R = gas constant (0.0821 L•atm/mol•K)
T = temperature of the gas (given in Kelvin)

First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15

Given T = 35 °C, converting to Kelvin:
T(K) = 35 + 273.15 = 308.15 K

Next, let's convert the given pressure from mmHg to atm:
1 atm = 760 mmHg
P(atm) = P(mmHg) / 760

Given P = 705 mmHg, converting to atm:
P(atm) = 705 mmHg / 760 = 0.927 atm

Now, let's substitute the values into the ideal gas law equation:

PV = nRT

V * 0.927 atm = n * 0.0821 L•atm/mol•K * 308.15 K

Since the number of moles of gas (n) is not given, we cannot solve for the volume directly. However, at STP, one mole of any gas occupies 22.4 liters. Therefore, we can assume that the number of moles of gas collected is equal to the volume of the gas at STP (V at STP).
So, V at STP = 22.4 liters

Now, let's solve for n using the volume given:

22.4 liters * 0.927 atm = n * 0.0821 L•atm/mol•K * 308.15 K

n = (22.4 liters * 0.927 atm) / (0.0821 L•atm/mol•K * 308.15 K)

n ≈ 0.997 moles

Since the number of moles and the volume at STP are equal, we can conclude that the volume of the gas at STP is approximately 0.997 liters.