Gas Laws Practice

A sample of C2H3Cl in a rigid container has its temperature changed from 483 K to 535 K. If its new pressure is 1.596 atm, what was its initial pressure in atmospheres?

• 1.44 atm

• 1.50 atm

• 1.60 atm

• 1.70 atm

Answer explanation

Using the ideal gas law, P1/T1 = P2/T2. Rearranging gives P1 = P2 * (T1/T2). Substituting P2 = 1.596 atm, T1 = 483 K, and T2 = 535 K results in P1 = 1.44 atm, confirming the correct choice.

The pressure of a 8.362 mol sample of C2H3Cl in a 72.87 L container is measured to be 16416 torr. What is the temperature of this gas in kelvins?

• 2294 K

• 2300 K

• 2400 K

• 2500 K

Answer explanation

Using the ideal gas law, PV = nRT, we can rearrange to find T = PV/nR. Substituting P = 16416 torr, V = 72.87 L, n = 8.362 mol, and R = 62.36 L·torr/(K·mol), we calculate T to be approximately 2294 K.

The temperature of a sample of COCl2 at a fixed volume is changed, causing a change in pressure from 54264 torr to 44764 torr. If its new temperature is 592.6 K, what was its original temperature in kelvins?

• 718.4 K

• 720 K

• 730 K

• 740 K

Answer explanation

Using the ideal gas law, P1/T1 = P2/T2. Rearranging gives T1 = P1*T2/P2. Substituting P1 = 54264 torr, P2 = 44764 torr, and T2 = 592.6 K results in T1 = 718.4 K, confirming the correct answer.

A sample of CH3Cl in a metal container at 10944 torr has its temperature changed from 1126.65 K to 252.9 K. What is the final pressure in torr?

• 2457 torr

• 2500 torr

• 2600 torr

• 2700 torr

Answer explanation

Using the ideal gas law, P1/T1 = P2/T2. Rearranging gives P2 = P1 * (T2/T1). Substituting P1 = 10944 torr, T1 = 1126.65 K, and T2 = 252.9 K results in P2 ≈ 2457 torr, confirming the correct answer.

A sample of ClF3 at 600 kPa experiences a change in volume from 7.24 L to 49.7 L. What is the resulting pressure in kilopascals?

• 90 kPa

• 100 kPa

• 110 kPa

• 120 kPa

Answer explanation

Using the ideal gas law, P1V1 = P2V2. Rearranging gives P2 = P1(V1/V2). Substituting P1 = 600 kPa, V1 = 7.24 L, and V2 = 49.7 L results in P2 = 90 kPa, confirming the correct answer.

A sample of ClF3 at 476.63 K experiences a change in volume from 82.9 L to 92.9 L. If its new pressure is 1.5743 atm at 743.6 K, what was its original pressure in atmospheres?

• 1.13 atm

• 1.20 atm

• 1.30 atm

• 1.40 atm

Answer explanation

Using the ideal gas law (PV=nRT), we can relate the initial and final states. Rearranging gives P1 = (P2 * V2 * T1) / (V1 * T2). Plugging in values, we find P1 = 1.13 atm, confirming the correct choice.

Tags

NGSS.HS-PS3-4

The temperature of 43.9 L of ClO2 is changed from 269 K to 798.7 K. During this, the pressure also changes from 537.6 kPa to 563 kPa. What is the new volume of the gas in liters?

• 124 L

• 130 L

• 140 L

• 150 L

Answer explanation

Using the combined gas law (P1V1/T1 = P2V2/T2), we can solve for the new volume V2. Plugging in the values: (537.6 kPa * 43.9 L / 269 K) = (563 kPa * V2 / 798.7 K). This calculation gives V2 = 124 L.

The temperature of a 33.967 L sample of N2O is changed, causing a change in pressure from 28576 torr to 52806.32 torr. If its new temperature is 1182.6 K and its new volume is 60.22 L, what was its original temperature in kelvins?

• 361.0 K

• 370.0 K

• 380.0 K

• 390.0 K

Answer explanation

Using the ideal gas law (PV=nRT), we can rearrange to find the original temperature. Plugging in the values for pressure, volume, and the new conditions, we calculate the original temperature to be 361.0 K.

A sample of C2H3Cl in a rigid container has its temperature changed from 483 K to 535 K. If its new pressure is 1.596 atm, what was its initial pressure in atmospheres?

• 1.44 atm

• 1.50 atm

• 1.60 atm

• 1.70 atm

Answer explanation

Using the ideal gas law, P1/T1 = P2/T2. Rearranging gives P1 = P2 * (T1/T2). Substituting P2 = 1.596 atm, T1 = 483 K, and T2 = 535 K results in P1 = 1.44 atm, confirming the correct choice.

A sample of TeF6 experiences a change in volume from 40.7 L to 90.98 L. If its new pressure is 372.5 kPa, what was its original pressure in kilopascals?

• 833 kPa

• 850 kPa

• 870 kPa

• 890 kPa

Answer explanation

Using the ideal gas law (P1V1 = P2V2), we can rearrange to find P1. Substituting V1 = 40.7 L, V2 = 90.98 L, and P2 = 372.5 kPa gives P1 = (372.5 kPa * 90.98 L) / 40.7 L = 833 kPa.