Question

A rigid cylinder contains a sample of gas at STP. What is the pressure of this gas after the sample is heated to 410 K?
(1) 1.0 atm (3) 0.67 atm
(2) 0.50 atm (4) 1.5 atm

Answer 1

To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the conditions of the gas sample, we calculate the pressure as follows: (NOTE: At STP 1 mol is equal to 22.4 L of a molecule)

P = nRT / V
P = (1 mol) (8.314 Pa m^3/ mol K) (298.15 K) / .0224 L
P = 110660 Pa or 1.0 atm

Thus, the answer is the value in option 1.

Answer 2

Answer:(4) 1.5 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas at STP= 1 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas at STP = V L

$V_2$ = final volume of gas = V L

$T_1$ = initial temperature of gas at STP= 273K

$T_2$ = final temperature of gas = 410 K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{1atm\times VL}{273K}=\frac{P_2\times VL}{410K}$

$P_2=1.50atm$

Therefore, the pressure of gas will be 1.50 atm.