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What is a expression of Charles's law?
Charles's law states that the volume of a gas is directly proportional to its temperature, at constant pressure. Mathematically, it can be expressed as: V1/T1 = V2/T2, where V represents volume and T represents temperature.
What expression of Charles's law?
Charles's law states that the volume of a gas is directly proportional to its temperature, at constant pressure. Mathematically, V1/T1 = V2/T2, where V is volume and T is temperature. This means that as the temperature of a gas increases, its volume also increases proportionally.
Which of the following is not a valid equation for describing the behavior for gases?
(p1/v1) = (p2/v2)For Apex (P1 N1)= (P2N2 )
How do you solve boyles law equation for V2?
To solve Boyle's Law equation for V2, first write the equation as P1V1 = P2V2. Then rearrange it to isolate V2 on one side, dividing both sides by P2 to solve for V2, which will be V2 = (P1 * V1) / P2.
Can i know more Details of position Vector?
The position vector is four dimensional, a quaternion: p= w +ix + jy + kz = w = v where w=ct is a real number and v is a three dimensional vector v=ix +jy +kz. The product of positions is a transformation: p1p2 = (w1 +v1)(w2 +v2)= (w1w2 -v1.v2) + (w1v2 + w2v1 + v1xv2) The '.' and 'x' are the cosine ans sine of the angles between v1 and v2 thus: p1p2= (w1 +v1)(w2 +v2)= (w1w2 -v1v2cos(v1v2)) + (w1v2 + w2v1 + v1v2sin(v1v2)) If v1 is perpendicular to v2 then cos(v1v2) is zero and sin(v1v2) is +- 1. if v1 is parallel to v2 then sin(v1v2) is zero and cos(v1v2) is +-1. Einstein's Relativity "interval" is a contraction of p1p2= ct1ct2 - x1x2 -y1y2 -z1z2. p1p2 is also |p1p2|( Cos(p1 +p2) + sin(p1 +p2)). This implies that Einstein's "interval" is cos(p1 +p2) and the interval is zero when cos(p1 +p2) is zero or the sum of the angles p1 +p2 is a multiple of 90 degrees.
What must the pressure and volume be if the pressure of its container is reduced to half its original size?
initial volume = V1 final volume = V2 initial pressure = P1 final pressure = P2 = (1/2)P1 P1V1 = P2V2 P1V1 = (1/2)P1V2 P1 cancels; V1 = (1/2)V2 V2 = 2V2.
If you initially have a gas at a pressure of 12 ATM a volume of 23 liters and then you raise the pressure to 14 ATM what is the new volume of the gas?
The volume of the gas will decrease proportionally to the increase in pressure, following Boyle's Law. Using the formula P1V1 = P2V2, where P1 = 12 ATM, V1 = 23 L, and P2 = 14 ATM, we can solve for V2 to find the new volume of the gas. Solving for V2 gives V2 = (P1)(V1) / P2 = (12)(23) / 14 = 19.71 liters.
What is p2 in boyles law?
In Boyle's Law, p2 represents the final pressure when a gas undergoes a change in volume at constant temperature. The law states that the initial pressure (p1) times the initial volume (V1) is equal to the final pressure (p2) times the final volume (V2), where p1V1 = p2V2.
What happens to the air pressure inside a balloon when the balloon is squeezed to half its volume at constant temperature?
Assuming the temperature is constant and none of the air is let out of the balloon, if you half the volume, the air pressure will double. If you want it, here is the math: Air is approximately an ideal gas, so when the temperature is constant, it follows the gas law of P1*V1 = P2*V2 where P1 is the pressure at state 1 (i.e. before the balloon is squeezed), V1 is the volume at state 1, P2 is the pressure at state 2 (i.e. after the balloon is squeezed), and V2 is the volume at state 2. When you half the volume, mathematically saying V2 = 0.5*V1. When you substitute this in, your equation now becomes P1*V1 = P2*0.5*V1. The V1 on both sides will cancel, leaving you with P1 = 0.5*P2. Since you are asking about the pressure after you squeeze the balloon, you want to know P2. So if you divide both sides by 0.5, you now have P1/0.5 = P2 or equivalently P2 = 2*P1 and we see that the pressure is twice that of the original.
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
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 )
The pressure of x mL of a gas was increased from 2 atm to 3 atm at constant temperature what will be the new volume?
If the pressure of a gas increases from 2 atm to 3 atm at constant temperature, the new volume will decrease by a factor equal to the initial pressure divided by the final pressure (V2 = V1 * P1/P2). So, the new volume would be 2/3 times the initial volume.
A sample of helium gas at 25 celsius is compressed from 200mL to 0.240 mL Its pressure is now 300mm Hg What was the original pressure of the helium?
To solve for the original pressure of the helium gas, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Using this law, we can set up the equation (P1)(V1) = (P2)(V2), where P1 is the original pressure, V1 is the original volume, P2 is the final pressure, and V2 is the final volume. Plugging in the values gives us (P1)(200 mL) = (300 mm Hg)(0.240 mL). Solving for P1 gives us P1 = (300 mm Hg)(0.240 mL) / 200 mL = 0.36 mm Hg. Therefore, the original pressure of the helium gas was 0.36 mm Hg.
At standard temperature a gas has a volume of 350 mL the temperature is then increased to 137 C and the pressure is held constant What is the new volume?
T1 = 273.15K. T2 = 410.15K. V1 = 350mL. V2 = ? P1 = P2. Since pressure is constant you can use the formula. V1/T1 = V2/T2 Rearrange the formula to get: V2 = T2V1/T1
