Chapter 13, Problem 3STP

Interpretation Introduction

Interpretation:

The new pressure of the gas have to be calculated.

Concept Introduction:

The combined gas law can be expressed as-

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

The equation is derived from the ideal gas law when amount of gas remains constant.

Initial pressure = P₁ Final pressure = P₂

Initial volume = V₁ Final volume = V₂

Initial temperature = T₁ Final temperature = T₂

Answer to Problem 3STP

The new pressure of the gas is-

B. 2.70 atm

Explanation of Solution

Data given: P₁ = 1.00 atm, V₁ = 5.00 L and T₁ = 20 °C = 293 K

V₂ = 2.00 L, T₂ = 43 °C = 316 K

  $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

By putting all the values in the above equation, we can calculate P₂

  $\begin{array}{l}{P}_2=\frac{P_1{V}_1{T}_2}{V_2{T}_1}\\ =\frac{1.00atm\times 5.00L\times 316K}{2.00L\times 293K}\\ =2.696atm\\ \simeq 2.70atm\end{array}$

The new pressure of the gas is 2.70 atm which is option B.

Conclusion

Reasons for incorrect options:

From the calculation we got that the new pressure of the gas is 2.70 atm.

So the answer is option B.

That means other options that is,

A. 0.812 atm

C. 3.14 atm

D. 5.38 atm are found to be incorrect.