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If the volume of a gas collected at a temperature of 600 °C and pressure of 1.05 * 1.05 Nm is 60 dms, what would be the volume of gas at STP (P=1.01 * 10^3 Nm, T = 273 K)?

A) 25 cm
C) 100 cm
B) 75 cm
D) 51 cm​

1 Answer

3 votes

Final answer:

To find the volume of gas at STP, use the ideal gas law equation PV = nRT. Convert the initial temperature to Kelvin and calculate the number of moles using the initial volume, temperature, and pressure. Therefore, the volume of gas at STP would be approximately 197.17 cm.

Step-by-step explanation:

To find the volume of gas at STP (standard temperature and pressure), we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the initial temperature and pressure to Kelvin: 600 °C = 600 + 273 = 873 K and 1.05 * 1.05 Nm = 1.05 * 1.05 * 10^5 Pa = 110250 Pa. Next, we need to find the number of moles using the initial volume and these converted values:

n = PV / RT = (110250 Pa * 60 dms) / (8.314 J/(mol K) * 873 K) = 911.67 mol

Now, we can use the number of moles and the given pressure and temperature at STP to find the final volume:

V = nRT / P = (911.67 mol * 8.314 J/(mol K) * 273 K) / (1.01 * 10^3 N/m^2) = 19716.91 dms = 197.17 cm

Therefore, the volume of gas at STP would be approximately 197.17 cm.

User Mathias Kemeter
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