Lesya P.

asked • 03/21/23

IDEAL GAS LAW EQUATION

0.465 MOLE SAMPLE OF GAS EXERTS 4.43 ATM OF PRESSURE IN 2.25 L CONTAINER. WHAT IS TEMPERATURE OF THE GAS IN THE CONTAINER?

A GAS HAS PRESSURE OF 0.410 ATM A VALUME 32L AT WHAT PRESSURE WOULD THE VALUME OF GAS CHANGE TO 28L?

WHAT THE VALUME OF GAS AT 50.0 C AND 0.80 ATM, IF THERE ARE 0.75 MOLES OF THE GAS PRESENT?

Jill K.

The ideal gas equation is PV = nRT. This can be written out as Pressure x Volume = moles of gas x R (gas constant) x T (temperature). The version of R to use here is R=0.08206 Liters x Atmospheres/Moles of gas x Temperature in K. This will keep all your units consistent. Try plugging in the numbers to the equation and see what you get. Contact me if you need more help. Thanks, Jill
Report

03/29/23

Zahid R.

tutor
Hello! Let's take a look at each of your questions one by one. For the first question, we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin. Rearranging this equation to solve for temperature, we get T = PV/nR. Plugging in the values given, we get: T = (4.43 atm)(2.25 L)/(0.465 mol)(0.08206 L∙atm/K∙mol) = 271 K Therefore, the temperature of the gas in the container is 271 Kelvin. For the second question, we can use the combined gas law, which states that P1V1/T1 = P2V2/T2, where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2 and V2 are the final pressure and volume, respectively. Rearranging this equation to solve for P2, we get P2 = (P1V1T2)/(V2T1). Plugging in the values given, we get: P2 = (0.410 atm)(32 L)(273 K)/(28 L)(298 K) = 0.496 atm Therefore, the pressure of the gas would need to increase to 0.496 atm in order for the volume to decrease to 28 L. For the third question, we can use the ideal gas law again to solve for volume. Plugging in the given values and solving for V, we get: V = nRT/P = (0.75 mol)(0.08206 L∙atm/K∙mol)(323 K)/(0.80 atm) = 19.3 L Therefore, the volume of the gas at 50.0 C and 0.80 atm with 0.75 moles of gas present would be 19.3 liters.
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03/30/23

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