Question

A 760 mL gas sample at STP is compressed

to a volume of 207 mL, and the temperature is
increased to 33°C. What is the new pressure
of the gas in Pa?
Answer in units of Pa. Answer in units of
Pa.

Answer 1

We can use the combined gas law to solve this problem:

(P1 x V1) / (T1) = (P2 x V2) / (T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

At STP (Standard Temperature and Pressure), the conditions are:

P1 = 101.325 kPa

V1 = 760 mL = 0.76 L

T1 = 273 K

After the gas is compressed, the new conditions are:

V2 = 207 mL = 0.207 L

T2 = 33°C + 273 = 306 K

We can plug in these values and solve for P2:

(101.325 kPa x 0.76 L) / 273 K = (P2 x 0.207 L) / 306 K

P2 = (101.325 kPa x 0.76 L x 306 K) / (0.207 L x 273 K)

P2 = 663.9 kPa

Therefore, the new pressure of the gas in Pa is 663.9 kPa, which is equivalent to 6.64 x 10^5 Pa.