isothermal compression of air in a slow-speed compressor Given gas constant (R) = 254.0 J/kg*K; gas temperature (T) = 284.0 K; final pressure (p_2) = 520.0 kPa; initial pressure (p_1) = 145.0 kPa, determine the specific work (w) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except w is given, so isolate w symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that w stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 254.0 J/kg*K, gas temperature (T) = 284.0 K, final pressure (p_2) = 520.0 kPa, initial pressure (p_1) = 145.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w = 92.1245\ \text{kJ/kg}$$

6. Step 6 — Check: returning w = 92.1245 kJ/kg to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 184.2 — kept a factor of two that cancels in the correct rearrangement.
• 46.0623 — dropped that same factor in the other direction.
• 101.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression of natural gas held at constant temperature Given gas constant (R) = 251.0 J/kg*K; final pressure (p_2) = 770.0 kPa; initial pressure (p_1) = 121.0 kPa; specific work (w) = 133.0 kJ/kg, determine the gas temperature (T) in K.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that T stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 251.0 J/kg*K, final pressure (p_2) = 770.0 kPa, initial pressure (p_1) = 121.0 kPa, specific work (w) = 133.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$T = 286.3\ \text{K}$$

6. Step 6 — Check: returning T = 286.3 K to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 572.7 — kept a factor of two that cancels in the correct rearrangement.
• 143.2 — dropped that same factor in the other direction.
• 315.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression work for a gas storage cylinder Given gas constant (R) = 254.0 J/kg*K; gas temperature (T) = 307.0 K; initial pressure (p_1) = 104.0 kPa; specific work (w) = 143.0 kJ/kg, determine the final pressure (p_2) in kPa.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 254.0 J/kg*K, gas temperature (T) = 307.0 K, initial pressure (p_1) = 104.0 kPa, specific work (w) = 143.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p_{2} = 650.8\ \text{kPa}$$

6. Step 6 — Check: returning p_2 = 650.8 kPa to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,302 — kept a factor of two that cancels in the correct rearrangement.
• 325.4 — dropped that same factor in the other direction.
• 715.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression of air in a slow-speed compressor Given gas constant (R) = 277.0 J/kg*K; gas temperature (T) = 293.0 K; final pressure (p_2) = 710.0 kPa; initial pressure (p_1) = 137.0 kPa, determine the specific work (w) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except w is given, so isolate w symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that w stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 277.0 J/kg*K, gas temperature (T) = 293.0 K, final pressure (p_2) = 710.0 kPa, initial pressure (p_1) = 137.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w = 133.5\ \text{kJ/kg}$$

6. Step 6 — Check: returning w = 133.5 kJ/kg to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 267.1 — kept a factor of two that cancels in the correct rearrangement.
• 66.7664 — dropped that same factor in the other direction.
• 146.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression of natural gas held at constant temperature Given gas constant (R) = 279.0 J/kg*K; final pressure (p_2) = 430.0 kPa; initial pressure (p_1) = 148.0 kPa; specific work (w) = 60.0000 kJ/kg, determine the gas temperature (T) in K.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that T stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 279.0 J/kg*K, final pressure (p_2) = 430.0 kPa, initial pressure (p_1) = 148.0 kPa, specific work (w) = 60.0000 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$T = 201.6\ \text{K}$$

6. Step 6 — Check: returning T = 201.6 K to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 403.3 — kept a factor of two that cancels in the correct rearrangement.
• 100.8 — dropped that same factor in the other direction.
• 221.8 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression work for a gas storage cylinder Given gas constant (R) = 270.0 J/kg*K; gas temperature (T) = 305.0 K; initial pressure (p_1) = 139.0 kPa; specific work (w) = 121.0 kJ/kg, determine the final pressure (p_2) in kPa.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 270.0 J/kg*K, gas temperature (T) = 305.0 K, initial pressure (p_1) = 139.0 kPa, specific work (w) = 121.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p_{2} = 604.1\ \text{kPa}$$

6. Step 6 — Check: returning p_2 = 604.1 kPa to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1,208 — kept a factor of two that cancels in the correct rearrangement.
• 302.1 — dropped that same factor in the other direction.
• 664.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression of air in a slow-speed compressor Given gas constant (R) = 291.0 J/kg*K; gas temperature (T) = 299.0 K; final pressure (p_2) = 690.0 kPa; initial pressure (p_1) = 128.0 kPa, determine the specific work (w) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except w is given, so isolate w symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that w stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 291.0 J/kg*K, gas temperature (T) = 299.0 K, final pressure (p_2) = 690.0 kPa, initial pressure (p_1) = 128.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w = 146.6\ \text{kJ/kg}$$

6. Step 6 — Check: returning w = 146.6 kJ/kg to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 293.2 — kept a factor of two that cancels in the correct rearrangement.
• 73.2903 — dropped that same factor in the other direction.
• 161.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression of natural gas held at constant temperature Given gas constant (R) = 275.0 J/kg*K; final pressure (p_2) = 460.0 kPa; initial pressure (p_1) = 137.0 kPa; specific work (w) = 124.0 kJ/kg, determine the gas temperature (T) in K.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that T stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 275.0 J/kg*K, final pressure (p_2) = 460.0 kPa, initial pressure (p_1) = 137.0 kPa, specific work (w) = 124.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$T = 372.3\ \text{K}$$

6. Step 6 — Check: returning T = 372.3 K to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 744.5 — kept a factor of two that cancels in the correct rearrangement.
• 186.1 — dropped that same factor in the other direction.
• 409.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression work for a gas storage cylinder Given gas constant (R) = 288.0 J/kg*K; gas temperature (T) = 303.0 K; initial pressure (p_1) = 118.0 kPa; specific work (w) = 267.0 kJ/kg, determine the final pressure (p_2) in kPa.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except p_2 is given, so isolate p_2 symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that p_2 stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 288.0 J/kg*K, gas temperature (T) = 303.0 K, initial pressure (p_1) = 118.0 kPa, specific work (w) = 267.0 kJ/kg.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p_{2} = 2516\ \text{kPa}$$

6. Step 6 — Check: returning p_2 = 2,516 kPa to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 5,032 — kept a factor of two that cancels in the correct rearrangement.
• 1,258 — dropped that same factor in the other direction.
• 2,767 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression

isothermal compression of air in a slow-speed compressor Given gas constant (R) = 288.0 J/kg*K; gas temperature (T) = 312.0 K; final pressure (p_2) = 840.0 kPa; initial pressure (p_1) = 147.0 kPa, determine the specific work (w) in kJ/kg.

Start with the thinking

• The governing relation printed in this handbook section is Isothermal compression work.
• Everything except w is given, so isolate w symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Isothermal compression of a gas requires work computed from the logarithmic pressure ratio at constant temperature.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

2. Step 2 — Rearrange the relation so that w stands alone on the left-hand side.

3. Step 3 — List the givens: gas constant (R) = 288.0 J/kg*K, gas temperature (T) = 312.0 K, final pressure (p_2) = 840.0 kPa, initial pressure (p_1) = 147.0 kPa.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$w = 156.6\ \text{kJ/kg}$$

6. Step 6 — Check: returning w = 156.6 kJ/kg to

   $$w = R T \ln\left(\dfrac{p_2}{p_1}\right)$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 313.2 — kept a factor of two that cancels in the correct rearrangement.
• 78.3081 — dropped that same factor in the other direction.
• 172.3 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isothermal Compression