Question

Suppose 8.2 kg of steam at 2.4 atm pressure is trapped inside a closed container of fixed volume at a temperature of 430.0 K. A Bunsen burner applies 742,000 J of thermal energy to the steam. What is the pressure of the steam inside the container after heating? The specific heat of steam is 2,010 J/(kg·K).

A. 1.2 atm

B. 2.7 atm

C. 3.5 atm

D. 4.8 atm

Answer 1

Answer:

B

Explanation:

B. 2.7 atm

Answer 2

The pressure of the steam inside the container after heating = 2.7 atm

Option B is correct.

Explanation:

Mass of steam (m) = 8.2 kg

Initial pressure $P_{1}$ = 2.4 atm

Initial temperature $T_{1}$ = 430 K

Heat supplied Q = 742000 J

Specific heat of steam = 2010  $\frac{J}{Kg k}$

Now heat supplied to the steam is given by the formula

⇒ Q = m × C × ( $T_{2}$ - $T_{1}$ )

⇒ Q = 8.2 × 2010 × ($T_{2}$ - 430 )

⇒ 742000 =  8.2 × 2010 × ($T_{2}$ - 430 )

⇒ $T_{2}$ = 475 K

This is the value of final temperature.

Since volume of the steam is constant so the pressure of the steam is directly proportional to the temperature of steam.

⇒ P ∝ T

⇒ $\frac{P_{2} }{P_{1} }$ = $\frac{T_{2} }{T_{1} }$

⇒ $P_{2}$ = $\frac{T_{2} }{T_{1} }$ × $P_{1}$

⇒ $P_{2}$ = $\frac{475}{430}$ × 2.4

⇒ $P_{2}$ = 2.65 ≈ 2.7 atm

The pressure of the steam inside the container after heating = 2.7 atm