Question

Using the first volume and temperature reading on the table as V1 and T1, solve for the unknown values in the table below. Remember to use the rules of significant figures when entering your numeric response.

Answer 1

Answer: Th value of A is $3.8\times 10^2K$, value of B is 0.84 L, value of C is 1.1 L and the value of D is $2.2\times 10^2K$

Explanation:

To calculate the missing values of volume and temperature, we will use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas at constant number of moles and pressure. Equation given by this law is:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$      ....(1)

where,

$V_1\text{ and }T_1$ are initial volume and temperature of the gas

$V_2\text{ and }T_2$ are final volume and temperature of the gas

• For A:

$V_1=1.0L\\T_1=295K\\V_2=1.3L\\T_2=AK$

Putting values in equation 1, we get:

$\frac{1.0L}{295K}=\frac{1.3L}{A}\\\\A=383.5K\approx 3.8\times 10^2K$

Whenever multiplication and division are involved, the answer must not contain, more number of significant figures as there are in the least precise term.

Hence, the value of A is $3.8\times 10^2K$

• For B:

$V_1=1.0L\\T_1=295K\\V_2=BL\\T_2=250K$

Putting values in equation 1, we get:

$\frac{1.0L}{295K}=\frac{B}{250K}\\\\B=0.84L$

Hence, the value of B is 0.84L

• For C:

$V_1=1.0L\\T_1=295K\\V_2=CL\\T_2=325K$

Putting values in equation 1, we get:

$\frac{1.0L}{295K}=\frac{C}{325K}\\\\C=1.1L$

Hence, the value of C is 1.1L

• For D:

$V_1=1.0L\\T_1=295K\\V_2=0.75L\\T_2=DK$

Putting values in equation 1, we get:

$\frac{1.0L}{295K}=\frac{0.75L}{D}\\\\D=221.25\approx 2.2\times 10^2K$

Hence, the value of D is $2.2\times 10^2K$

Answer 2

Answer:

A= 384

B= 0.85

C= 1.1

D= 221