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What happens to the pressure of a gas when the volume of the container double in size?
You have for an Ideal Gas:PV = mRT/M( P2 ) ( V2 )/ (T2 ) ( m2 ) = ( P1 ) ( V1 ) / ( m1 ) ( T1 ) = R/M = ConstantV2 = ( V1 ) ( P1 /P2 ) ( T2/T1 ) ( m2 /m1 )You have :( P1 / P2 ) = 1.00( T2 / T1 ) = 1.00( m2 / m1 ) = 2.00V2 = ( V1 ) ( 1.000 ) ( 1.000 ( 2.000 ) = ( 2.000 ) ( V1 )
What is the final pressure of a 2.00 ATM gas container that increased in temperature from 299 K to 333 K while the volume increased from 650 mL to 850 mL?
P1 = 2T1 = 299T2 = 333V1 = 0.65V2 = 0.85P2t = P1 * T2 / T1 = 2.227 ATMP2v = P1 * V1 / V2 = 1.529 ATMP2 = P2t * P2v = 3.41 ATM
The gas pressure in an aerosol can is 1.8 ATM at 25oC If the gas is an ideal gas what pressure would develop in the can if it were heated to 475oC?
Using the ideal gas law (PV = nRT), we can calculate the new pressure of the gas in the aerosol can. Given that the initial pressure (P1) is 1.8 ATM and the initial temperature (T1) is 25°C, we can rearrange the formula to find the new pressure (P2) at 475°C. Since the volume (V), moles of gas (n), and gas constant (R) remain constant, we can solve for P2: P2 = (P1 * T2) / T1 = (1.8 ATM * 748 K) / 298 K ≈ 4.5 ATM.
Derive the formulaof gay-lussac's law?
Gay-Lussac's Law states that the pressure of a fixed amount of gas is directly proportional to its temperature, assuming constant volume and amount of gas. The formula is expressed as P1/T1 = P2/T2, where P represents pressure and T represents temperature. This law is applicable only when the volume and quantity of gas are held constant.
A sample of oxygen gas has a volume of 150 milliliters at 300k if the pressure of the sample is held constant and the temp raised to 600k the new sample will be?
Using the combined gas law (P1V1/T1 = P2V2/T2), we can calculate the new volume of the oxygen gas sample at 600K. Given P1V1/T1 = P2V2/T2, we have P1 = P2 (pressure is constant), V1 = 150 mL, T1 = 300K, and T2 = 600K. Plugging in these values, we get V2 = (P1 * V1 * T2) / (T1) = (1 * 150 * 600) / (300) = 300 mL. So, the new volume of the oxygen gas sample at 600K would be 300 milliliters.
What is the formula of gay-lussac's law when the T2 is missing?
Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature at constant volume. The formula is P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 is the final pressure.
What is the formula for gay lussacs law?
Gay-Lussac's Law states that the pressure of a sample of gas at constant volume, is directly proportional to its temperature in Kelvin. The P's represent pressure, while the T's represent temperature in Kelvin. P1 / T1 = constant After the change in pressure and temperature, P2 / T2 = constant Combine the two equations: P1 / T1 = P2 / T2 When any three of the four quantities in the equation are known, the fourth can be calculated. For example, we've known P1, T1 and P2, the T2 can be: T2 = P2 x T1 / P1
What is gay lussacs law formula?
Gay-Lussac's Law states that the pressure of a sample of gas at constant volume, is directly proportional to its temperature in Kelvin. The P's represent pressure, while the T's represent temperature in Kelvin. P1 / T1 = constant After the change in pressure and temperature, P2 / T2 = constant Combine the two equations: P1 / T1 = P2 / T2 When any three of the four quantities in the equation are known, the fourth can be calculated. For example, we've known P1, T1 and P2, the T2 can be: T2 = P2 x T1 / P1
What is the formula for Gay-Lussac?
Gay-Lussac's Law states that the pressure of a sample of gas at constant volume, is directly proportional to its temperature in Kelvin. The P's represent pressure, while the T's represent temperature in Kelvin. P1 / T1 = constant After the change in pressure and temperature, P2 / T2 = constant Combine the two equations: P1 / T1 = P2 / T2 When any three of the four quantities in the equation are known, the fourth can be calculated. For example, we've known P1, T1 and P2, the T2 can be: T2 = P2 x T1 / P1
What is law in connection to pressure and temperature?
Gay-Lussac's law. P1/T1 = P2/T2
How can one solve Gay-Lussac's Law?
To solve Gay-Lussac's Law, use the formula P1/T1 P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature. Rearrange the formula to solve for the unknown variable.
How do you write the addition problem to If one plant can capture one insect in one second how many insects could 25 plants trap in 2 seconds?
Let P = Plant Let T = Time Let I = Insect P1 x T1 / I1 = P2 x T2 / I2 I2 =(P2)(T2)(I1) / (P1)(T1) P1 = 1 T1 = 1 I1 = 1 P2 = 25 T2 = 2 I2 = number of insects = 50
If a gas has a pressure of 50.0 mm Hg at 540 K what will be the pressure at 200K if the volume does not change?
If you dcrease the temperature you will decrease the pressure proportionately. So, T1 over T2 will equal P1 over P2. We can derive the formula P1 x T2 = P2 x T1. Substitue the values and we get 50.0 mm x 200K = P2 x 540K 10,000 mmK = P2 x 540K 10,000mmK / 540K = P2 P2 = 18.52 mm of Mercury in a constant volume
How do you find T2 of the combined gas law?
The Combined Gas Law relates pressure (P), volume (V) and temperature (T). The appropriate SI units are P in atm, V in liters, and T in degrees Kelvin. The Combined Gas Law equation is (P1*V1)/T1 = (P2V2)/T2. Isolating for V2 the equation then becomes (P1V1T2)/(T1P2) = V2
How do you calculate water vapor pressure?
Water vapor pressure can be calculated using the Clausius-Clapeyron equation, which relates vapor pressure to temperature. The equation is: ln(P2/P1) (Hvap/R)(1/T1 - 1/T2), where P1 and P2 are the vapor pressures at temperatures T1 and T2, Hvap is the heat of vaporization, and R is the gas constant.
What is the relationship between density and temperature of CNG?
At low pressures you can use the ideal gas equation: (P1*V1)/T1 = (P2*V2)/T2 At constant volume, the equation will be: P1/T1 = P2/T2 At higher pressures (appr. above 10 bar) the deviation to real gas becomes significant, hence the compression factor (Z) is introduced.
What happens to the pressure of a gas when the volume of the container double in size?
You have for an Ideal Gas:PV = mRT/M( P2 ) ( V2 )/ (T2 ) ( m2 ) = ( P1 ) ( V1 ) / ( m1 ) ( T1 ) = R/M = ConstantV2 = ( V1 ) ( P1 /P2 ) ( T2/T1 ) ( m2 /m1 )You have :( P1 / P2 ) = 1.00( T2 / T1 ) = 1.00( m2 / m1 ) = 2.00V2 = ( V1 ) ( 1.000 ) ( 1.000 ( 2.000 ) = ( 2.000 ) ( V1 )
