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STP Temperature and Pressure Calculator

Standard temperature and pressure, molar volume and gas law conversions with step-by-step solutions
Find the volume of 2.50 mol of an ideal gas at STP
Give the temperature and pressure at STP in kelvin and kilopascals
A gas occupies 4.50 L at 30.0 C and 1.25 atm. What is its volume at STP?
How many moles are in 11.2 L of an ideal gas at STP?

The STP Numbers, and Which Set to Quote

Standard temperature and pressure (STP) is an agreed reference state, not a law of nature. It exists so that gas volumes measured in different labs can be compared.

Current IUPAC definition:

T=273.15 K=0 ∘C,P=105 Pa=100 kPa=1 barT = 273.15\ \text{K} = 0\ ^\circ\text{C}, \qquad P = 10^{5}\ \text{Pa} = 100\ \text{kPa} = 1\ \text{bar}

Older definition, still the one used in most textbooks and on school exams:

T=273.15 K,P=1 atm=101.325 kPa=760 mmHgT = 273.15\ \text{K}, \qquad P = 1\ \text{atm} = 101.325\ \text{kPa} = 760\ \text{mmHg}

The molar volume of an ideal gas follows from whichever you pick:

  • 1 atm convention: Vm=22.414V_m = 22.414 L/mol
  • 1 bar convention: Vm=22.711V_m = 22.711 L/mol

STP is not SATP. Standard ambient temperature and pressure is 298.15 K (25 °C) at 100 kPa, giving Vm=24.79V_m = 24.79 L/mol.

The assumption people forget: the 22.4 L/mol shortcut is tied to 0 °C and 1 atm specifically. Quote the convention you are using, because the two answers differ by 1.3%.

Using STP in the Gas Laws

Every STP calculation is the ideal gas law with the temperature and pressure already fixed:

PV=nRTPV = nRT

  • PP — pressure, pascals (Pa) with the SI gas constant, or atmospheres (atm) with the L·atm form
  • VV — volume, cubic metres (m³) or litres (L)
  • nn — amount of substance, moles (mol)
  • TT — absolute temperature, kelvin (K)
  • RR — gas constant, 8.314 J/(mol·K) or 0.08206 L·atm/(mol·K)

Substituting the STP values for one mole gives the molar volume directly:

Vm=RTP=(0.08206)(273.15)1.00=22.414 L/molV_m = \frac{RT}{P} = \frac{(0.08206)(273.15)}{1.00} = 22.414\ \text{L/mol}

To move a sample measured elsewhere to STP, use the combined gas law:

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

When it applies: a fixed amount of gas, well above its condensation point and below a few atmospheres. Near liquefaction, real gases deviate and you need a compressibility factor ZZ in PV=ZnRTPV = ZnRT.

Common Mistakes to Avoid

  • Using celsius in the gas law — every TT here is absolute. Add 273.15 first; a temperature of 0 in the denominator is the classic tell.
  • Mixing RR with the wrong pressure unit — 0.08206 goes with atm and litres, 8.314 goes with pascals and cubic metres. Pick one system and stay in it.
  • Quoting 22.4 L/mol under the bar convention — at 100 kPa the molar volume is 22.711 L/mol.
  • Confusing STP with room temperature — STP is 0 °C, which is why gas volumes at STP look smaller than lab measurements.
  • Applying the molar volume to a liquid or solid — it is a property of the ideal gas model only.
  • Forgetting that nn must be constant in the combined gas law. If gas is added or consumed by reaction, go back to PV=nRTPV = nRT on each side.

示例题目

Step 1: V=nRT/PV = nRT/P with R=0.08206 L\cdotpatm/(mol\cdotpK)R = 0.08206\ \text{L·atm/(mol·K)}, T=273.15 KT = 273.15\ \text{K}, P=1.00 atmP = 1.00\ \text{atm}
Step 2: RT=(0.08206 L\cdotpatm/(mol\cdotpK))(273.15 K)=22.414 L\cdotpatm/molRT = (0.08206\ \text{L·atm/(mol·K)})(273.15\ \text{K}) = 22.414\ \text{L·atm/mol}
Step 3: V=(2.50 mol)(22.414 L\cdotpatm/mol)÷(1.00 atm)V = (2.50\ \text{mol})(22.414\ \text{L·atm/mol}) \div (1.00\ \text{atm})
Step 4: V=56.04 LV = 56.04\ \text{L}
Answer: V≈56.0V \approx 56.0 L

Step 1: Vm=RT/PV_m = RT/P with R=8.314 J/(mol\cdotpK)R = 8.314\ \text{J/(mol·K)} and P=1.00×105 PaP = 1.00 \times 10^{5}\ \text{Pa}
Step 2: RT=(8.314 J/(mol\cdotpK))(273.15 K)=2270.98 J/molRT = (8.314\ \text{J/(mol·K)})(273.15\ \text{K}) = 2270.98\ \text{J/mol}
Step 3: Vm=(2270.98 J/mol)÷(1.00×105 Pa)=2.271×10−2 m3/molV_m = (2270.98\ \text{J/mol}) \div (1.00 \times 10^{5}\ \text{Pa}) = 2.271 \times 10^{-2}\ \text{m}^3\text{/mol}
Step 4: Convert: 2.271×10−2 m3=22.71 L2.271 \times 10^{-2}\ \text{m}^3 = 22.71\ \text{L}
Answer: Vm=22.71V_m = 22.71 L/mol (versus 22.4122.41 L/mol at 1 atm)

Step 1: Convert temperatures: T1=30.0+273.15=303.15 KT_1 = 30.0 + 273.15 = 303.15\ \text{K}, T2=273.15 KT_2 = 273.15\ \text{K}
Step 2: V2=P1V1T2T1P2=(1.25 atm)(4.50 L)(273.15 K)(303.15 K)(1.00 atm)V_2 = \dfrac{P_1 V_1 T_2}{T_1 P_2} = \dfrac{(1.25\ \text{atm})(4.50\ \text{L})(273.15\ \text{K})}{(303.15\ \text{K})(1.00\ \text{atm})}
Step 3: Numerator: 1.25×4.50×273.15=1536.5 atm\cdotpL\cdotpK1.25 \times 4.50 \times 273.15 = 1536.5\ \text{atm·L·K}
Step 4: V2=1536.5÷303.15=5.068 LV_2 = 1536.5 \div 303.15 = 5.068\ \text{L}
Step 5: n=V2/Vm=(5.068 L)÷(22.414 L/mol)=0.2261 moln = V_2 / V_m = (5.068\ \text{L}) \div (22.414\ \text{L/mol}) = 0.2261\ \text{mol}
Answer: V2≈5.07V_2 \approx 5.07 L at STP, which is 0.2260.226 mol

常见问题

Under the current IUPAC definition, STP is 273.15 K (0 °C) and 100 kPa (1 bar). Under the older definition still used in most textbooks it is 273.15 K and 1 atm = 101.325 kPa. Always state which one you used, because the molar volume differs: 22.71 L/mol versus 22.414 L/mol.

Standard temperature is 273.15 K exactly. That is the ice point, 0 °C. Many problems round it to 273 K, which changes a molar volume by less than 0.06 percent.

It is the ideal gas law PV = nRT with P and T pinned to the standard values. For one mole this collapses to the molar volume V = RT/P = 22.414 L/mol at 1 atm. To convert a sample measured at other conditions, use the combined gas law P₁V₁/T₁ = P₂V₂/T₂.

STP is 0 °C at 100 kPa (or 1 atm). SATP, sometimes written NTP, is 25 °C at 100 kPa and gives a molar volume of 24.79 L/mol. The gas is warmer at SATP, so the same amount of gas occupies about 9 percent more space.

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